Solve frac{(frac{2sqrt{3}}{3})^{2}+(2b)^{2}-frac{(sqrt{3})^2}{3}}{2*(frac{2sqrt{3}}{3})*(2b)}=frac{(frac{2sqrt{3}}{3})^2+b^2-(frac{2sqrt{3}}{3})^2}{2*(frac{2sqrt{3}}{3})b}

$$\frac { ( \frac { 2 \sqrt { 3 } } { 3 } ) ^ { 2 } + ( 2 b ) ^ { 2 } - \frac { ( \sqrt { 3 } ) ^ { 2 } } { 3 } } { 2 \times ( \frac { 2 \sqrt { 3 } } { 3 } ) \cdot ( 2 b ) } = \frac { ( \frac { 2 \sqrt { 3 } } { 3 } ) ^ { 2 } + b ^ { 2 } - ( \frac { 2 \sqrt { 3 } } { 3 } ) ^ { 2 } } { 2 \times ( \frac { 2 \sqrt { 3 } } { 3 } ) b }$$

Solve for b

$$b=-\frac{\sqrt{6}i}{6}\approx -0-0.40824829i$$

$$b=\frac{\sqrt{6}i}{6}\approx 0.40824829i$$

Steps by Finding Square Root

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+\left(2b\right)^{2}-\frac{\left(\sqrt{3}\right)^{2}}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Expand \left(2b\right)^{2}.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+2^{2}b^{2}-\frac{\left(\sqrt{3}\right)^{2}}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+4b^{2}-\frac{\left(\sqrt{3}\right)^{2}}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+4b^{2}-\frac{3}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Divide 3 by 3 to get 1.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+4b^{2}-1}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4b^{2}-1 times \frac{3^{2}}{3^{2}}.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+\frac{\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Since \frac{\left(2\sqrt{3}\right)^{2}}{3^{2}} and \frac{\left(4b^{2}-1\right)\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 2 and 2 to get 4.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{4\times \frac{2\sqrt{3}}{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Express 4\times \frac{2\sqrt{3}}{3} as a single fraction.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{\frac{4\times 2\sqrt{3}}{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Express \frac{4\times 2\sqrt{3}}{3}b as a single fraction.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{\frac{4\times 2\sqrt{3}b}{3}}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Divide \frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}} by \frac{4\times 2\sqrt{3}b}{3} by multiplying \frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}} by the reciprocal of \frac{4\times 2\sqrt{3}b}{3}.

$$\frac{\left(\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}\right)\times 3}{3^{2}\times 4\times 2\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Cancel out 3 in both numerator and denominator.

$$\frac{3^{2}\left(4b^{2}-1\right)+\left(2\sqrt{3}\right)^{2}}{2\times 3\times 4\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Factor the expressions that are not already factored in \frac{3^{2}\left(4b^{2}-1\right)+\left(2\sqrt{3}\right)^{2}}{2\times 3\times 4\sqrt{3}b}.

$$\frac{3\left(12b^{2}+1\right)}{2\times 3\times 4\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Cancel out 3 in both numerator and denominator.

$$\frac{12b^{2}+1}{2\times 4\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Rationalize the denominator of \frac{12b^{2}+1}{2\times 4\sqrt{3}b} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{2\times 4\left(\sqrt{3}\right)^{2}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

The square of \sqrt{3} is 3.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{2\times 4\times 3b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 2 and 4 to get 8.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{8\times 3b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 8 and 3 to get 24.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply b^{2} times \frac{3^{2}}{3^{2}}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+\frac{b^{2}\times 3^{2}}{3^{2}}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Since \frac{\left(2\sqrt{3}\right)^{2}}{3^{2}} and \frac{b^{2}\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{2^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{4\left(\sqrt{3}\right)^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

The square of \sqrt{3} is 3.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{4\times 3}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{12}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{12}{9}}{2\times \frac{2\sqrt{3}}{3}b}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{4}{3}}{2\times \frac{2\sqrt{3}}{3}b}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 3 is 9. Multiply \frac{4}{3} times \frac{3}{3}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{9}-\frac{4\times 3}{9}}{2\times \frac{2\sqrt{3}}{3}b}$$

Since \frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{9} and \frac{4\times 3}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9}}{2\times \frac{2\sqrt{3}}{3}b}$$

Express 2\times \frac{2\sqrt{3}}{3} as a single fraction.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9}}{\frac{2\times 2\sqrt{3}}{3}b}$$

Express \frac{2\times 2\sqrt{3}}{3}b as a single fraction.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9}}{\frac{2\times 2\sqrt{3}b}{3}}$$

Divide \frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9} by \frac{2\times 2\sqrt{3}b}{3} by multiplying \frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9} by the reciprocal of \frac{2\times 2\sqrt{3}b}{3}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\left(\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3\right)\times 3}{9\times 2\times 2\sqrt{3}b}$$

Cancel out 3 in both numerator and denominator.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{3^{2}b^{2}-3\times 4+\left(2\sqrt{3}\right)^{2}}{2\times 2\times 3\sqrt{3}b}$$

Factor the expressions that are not already factored in \frac{3^{2}b^{2}-3\times 4+\left(2\sqrt{3}\right)^{2}}{2\times 2\times 3\sqrt{3}b}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{9b^{2}}{3\sqrt{3}\times 2^{2}b}$$

Cancel out 3\sqrt{3} in both numerator and denominator.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\sqrt{3}b^{2}}{2^{2}b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\sqrt{3}b^{2}}{4b}$$

Use the distributive property to multiply 12b^{2}+1 by \sqrt{3}.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b}=\frac{\sqrt{3}b^{2}}{4b}$$

Subtract \frac{\sqrt{3}b^{2}}{4b} from both sides.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b}-\frac{\sqrt{3}b^{2}}{4b}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 24b and 4b is 24b. Multiply \frac{\sqrt{3}b^{2}}{4b} times \frac{6}{6}.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b}-\frac{6\sqrt{3}b^{2}}{24b}=0$$

Since \frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b} and \frac{6\sqrt{3}b^{2}}{24b} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}-6\sqrt{3}b^{2}}{24b}=0$$

Combine like terms in 12b^{2}\sqrt{3}+\sqrt{3}-6\sqrt{3}b^{2}.

$$\frac{\sqrt{3}+6b^{2}\sqrt{3}}{24b}=0$$

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 24b.

$$\sqrt{3}+6b^{2}\sqrt{3}=0$$

Subtract \sqrt{3} from both sides. Anything subtracted from zero gives its negation.

$$6b^{2}\sqrt{3}=-\sqrt{3}$$

Dividing by 6\sqrt{3} undoes the multiplication by 6\sqrt{3}.

$$b^{2}=-\frac{\sqrt{3}}{6\sqrt{3}}$$

Divide -\sqrt{3} by 6\sqrt{3}.

$$b^{2}=-\frac{1}{6}$$

Take the square root of both sides of the equation.

$$b=\frac{\sqrt{6}i}{6} b=-\frac{\sqrt{6}i}{6}$$

Steps Using the Quadratic Formula

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+\left(2b\right)^{2}-\frac{\left(\sqrt{3}\right)^{2}}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Expand \left(2b\right)^{2}.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+2^{2}b^{2}-\frac{\left(\sqrt{3}\right)^{2}}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+4b^{2}-\frac{\left(\sqrt{3}\right)^{2}}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+4b^{2}-\frac{3}{3}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Divide 3 by 3 to get 1.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+4b^{2}-1}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4b^{2}-1 times \frac{3^{2}}{3^{2}}.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+\frac{\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Since \frac{\left(2\sqrt{3}\right)^{2}}{3^{2}} and \frac{\left(4b^{2}-1\right)\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}\times 2b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 2 and 2 to get 4.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{4\times \frac{2\sqrt{3}}{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Express 4\times \frac{2\sqrt{3}}{3} as a single fraction.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{\frac{4\times 2\sqrt{3}}{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Express \frac{4\times 2\sqrt{3}}{3}b as a single fraction.

$$\frac{\frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}}}{\frac{4\times 2\sqrt{3}b}{3}}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Divide \frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}} by \frac{4\times 2\sqrt{3}b}{3} by multiplying \frac{\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}}{3^{2}} by the reciprocal of \frac{4\times 2\sqrt{3}b}{3}.

$$\frac{\left(\left(2\sqrt{3}\right)^{2}+\left(4b^{2}-1\right)\times 3^{2}\right)\times 3}{3^{2}\times 4\times 2\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Cancel out 3 in both numerator and denominator.

$$\frac{3^{2}\left(4b^{2}-1\right)+\left(2\sqrt{3}\right)^{2}}{2\times 3\times 4\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Factor the expressions that are not already factored in \frac{3^{2}\left(4b^{2}-1\right)+\left(2\sqrt{3}\right)^{2}}{2\times 3\times 4\sqrt{3}b}.

$$\frac{3\left(12b^{2}+1\right)}{2\times 3\times 4\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Cancel out 3 in both numerator and denominator.

$$\frac{12b^{2}+1}{2\times 4\sqrt{3}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Rationalize the denominator of \frac{12b^{2}+1}{2\times 4\sqrt{3}b} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{2\times 4\left(\sqrt{3}\right)^{2}b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

The square of \sqrt{3} is 3.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{2\times 4\times 3b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 2 and 4 to get 8.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{8\times 3b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 8 and 3 to get 24.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\left(\frac{2\sqrt{3}}{3}\right)^{2}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+b^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply b^{2} times \frac{3^{2}}{3^{2}}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}+\frac{b^{2}\times 3^{2}}{3^{2}}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

Since \frac{\left(2\sqrt{3}\right)^{2}}{3^{2}} and \frac{b^{2}\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}{2\times \frac{2\sqrt{3}}{3}b}$$

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{2^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{4\left(\sqrt{3}\right)^{2}}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

The square of \sqrt{3} is 3.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{4\times 3}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{12}{3^{2}}}{2\times \frac{2\sqrt{3}}{3}b}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{12}{9}}{2\times \frac{2\sqrt{3}}{3}b}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{3^{2}}-\frac{4}{3}}{2\times \frac{2\sqrt{3}}{3}b}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 3 is 9. Multiply \frac{4}{3} times \frac{3}{3}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{9}-\frac{4\times 3}{9}}{2\times \frac{2\sqrt{3}}{3}b}$$

Since \frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}}{9} and \frac{4\times 3}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9}}{2\times \frac{2\sqrt{3}}{3}b}$$

Express 2\times \frac{2\sqrt{3}}{3} as a single fraction.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9}}{\frac{2\times 2\sqrt{3}}{3}b}$$

Express \frac{2\times 2\sqrt{3}}{3}b as a single fraction.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9}}{\frac{2\times 2\sqrt{3}b}{3}}$$

Divide \frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9} by \frac{2\times 2\sqrt{3}b}{3} by multiplying \frac{\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3}{9} by the reciprocal of \frac{2\times 2\sqrt{3}b}{3}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\left(\left(2\sqrt{3}\right)^{2}+b^{2}\times 3^{2}-4\times 3\right)\times 3}{9\times 2\times 2\sqrt{3}b}$$

Cancel out 3 in both numerator and denominator.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{3^{2}b^{2}-3\times 4+\left(2\sqrt{3}\right)^{2}}{2\times 2\times 3\sqrt{3}b}$$

Factor the expressions that are not already factored in \frac{3^{2}b^{2}-3\times 4+\left(2\sqrt{3}\right)^{2}}{2\times 2\times 3\sqrt{3}b}.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{9b^{2}}{3\sqrt{3}\times 2^{2}b}$$

Cancel out 3\sqrt{3} in both numerator and denominator.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\sqrt{3}b^{2}}{2^{2}b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(12b^{2}+1\right)\sqrt{3}}{24b}=\frac{\sqrt{3}b^{2}}{4b}$$

Use the distributive property to multiply 12b^{2}+1 by \sqrt{3}.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b}=\frac{\sqrt{3}b^{2}}{4b}$$

Subtract \frac{\sqrt{3}b^{2}}{4b} from both sides.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b}-\frac{\sqrt{3}b^{2}}{4b}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 24b and 4b is 24b. Multiply \frac{\sqrt{3}b^{2}}{4b} times \frac{6}{6}.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b}-\frac{6\sqrt{3}b^{2}}{24b}=0$$

Since \frac{12b^{2}\sqrt{3}+\sqrt{3}}{24b} and \frac{6\sqrt{3}b^{2}}{24b} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12b^{2}\sqrt{3}+\sqrt{3}-6\sqrt{3}b^{2}}{24b}=0$$

Combine like terms in 12b^{2}\sqrt{3}+\sqrt{3}-6\sqrt{3}b^{2}.

$$\frac{\sqrt{3}+6b^{2}\sqrt{3}}{24b}=0$$

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 24b.

$$\sqrt{3}+6b^{2}\sqrt{3}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$6\sqrt{3}b^{2}+\sqrt{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6\sqrt{3} for a, 0 for b, and \sqrt{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times 6\sqrt{3}\sqrt{3}}}{2\times 6\sqrt{3}}$$

Square 0.

$$b=\frac{0±\sqrt{-4\times 6\sqrt{3}\sqrt{3}}}{2\times 6\sqrt{3}}$$

Multiply -4 times 6\sqrt{3}.

$$b=\frac{0±\sqrt{\left(-24\sqrt{3}\right)\sqrt{3}}}{2\times 6\sqrt{3}}$$

Multiply -24\sqrt{3} times \sqrt{3}.

$$b=\frac{0±\sqrt{-72}}{2\times 6\sqrt{3}}$$

Take the square root of -72.

$$b=\frac{0±6\sqrt{2}i}{2\times 6\sqrt{3}}$$

Multiply 2 times 6\sqrt{3}.

$$b=\frac{0±6\sqrt{2}i}{12\sqrt{3}}$$

Now solve the equation b=\frac{0±6\sqrt{2}i}{12\sqrt{3}} when ± is plus.

$$b=\frac{\sqrt{6}i}{6}$$

Now solve the equation b=\frac{0±6\sqrt{2}i}{12\sqrt{3}} when ± is minus.

$$b=-\frac{\sqrt{6}i}{6}$$

The equation is now solved.

$$b=\frac{\sqrt{6}i}{6} b=-\frac{\sqrt{6}i}{6}$$