Solve sqrt{(frac{2sqrt{93}}{3}+2)(2)(2)(frac{2sqrt{93}}{3}+2-4)}

$$\sqrt{ ( \frac{ 2 \sqrt{ 93 } }{ 3 } +2)(2)(2)( \frac{ 2 \sqrt{ 93 } }{ 3 } +2-4) }$$

Evaluate

$$\frac{8\sqrt{21}}{3}\approx 12.220201853$$

Short Solution Steps

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

$$\sqrt{\left(\frac{2\sqrt{93}}{3}+\frac{2\times 3}{3}\right)\times 2\times 2\left(\frac{2\sqrt{93}}{3}+2-4\right)}$$

Since \frac{2\sqrt{93}}{3} and \frac{2\times 3}{3} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{2\sqrt{93}+2\times 3}{3}\times 2\times 2\left(\frac{2\sqrt{93}}{3}+2-4\right)}$$

Do the multiplications in 2\sqrt{93}+2\times 3.

$$\sqrt{\frac{2\sqrt{93}+6}{3}\times 2\times 2\left(\frac{2\sqrt{93}}{3}+2-4\right)}$$

Multiply 2 and 2 to get 4.

$$\sqrt{\frac{2\sqrt{93}+6}{3}\times 4\left(\frac{2\sqrt{93}}{3}+2-4\right)}$$

Subtract 4 from 2 to get -2.

$$\sqrt{\frac{2\sqrt{93}+6}{3}\times 4\left(\frac{2\sqrt{93}}{3}-2\right)}$$

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

$$\sqrt{\frac{2\sqrt{93}+6}{3}\times 4\left(\frac{2\sqrt{93}}{3}-\frac{2\times 3}{3}\right)}$$

Since \frac{2\sqrt{93}}{3} and \frac{2\times 3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\sqrt{\frac{2\sqrt{93}+6}{3}\times 4\times \frac{2\sqrt{93}-2\times 3}{3}}$$

Do the multiplications in 2\sqrt{93}-2\times 3.

$$\sqrt{\frac{2\sqrt{93}+6}{3}\times 4\times \frac{2\sqrt{93}-6}{3}}$$

Express \frac{2\sqrt{93}+6}{3}\times 4 as a single fraction.

$$\sqrt{\frac{\left(2\sqrt{93}+6\right)\times 4}{3}\times \frac{2\sqrt{93}-6}{3}}$$

Multiply \frac{\left(2\sqrt{93}+6\right)\times 4}{3} times \frac{2\sqrt{93}-6}{3} by multiplying numerator times numerator and denominator times denominator.

$$\sqrt{\frac{\left(2\sqrt{93}+6\right)\times 4\left(2\sqrt{93}-6\right)}{3\times 3}}$$

Multiply 3 and 3 to get 9.

$$\sqrt{\frac{\left(2\sqrt{93}+6\right)\times 4\left(2\sqrt{93}-6\right)}{9}}$$

Use the distributive property to multiply 2\sqrt{93}+6 by 4.

$$\sqrt{\frac{\left(8\sqrt{93}+24\right)\left(2\sqrt{93}-6\right)}{9}}$$

Apply the distributive property by multiplying each term of 8\sqrt{93}+24 by each term of 2\sqrt{93}-6.

$$\sqrt{\frac{16\left(\sqrt{93}\right)^{2}-48\sqrt{93}+48\sqrt{93}-144}{9}}$$

The square of \sqrt{93} is 93.

$$\sqrt{\frac{16\times 93-48\sqrt{93}+48\sqrt{93}-144}{9}}$$

Multiply 16 and 93 to get 1488.

$$\sqrt{\frac{1488-48\sqrt{93}+48\sqrt{93}-144}{9}}$$

Combine -48\sqrt{93} and 48\sqrt{93} to get 0.

$$\sqrt{\frac{1488-144}{9}}$$

Subtract 144 from 1488 to get 1344.

$$\sqrt{\frac{1344}{9}}$$

Reduce the fraction \frac{1344}{9} to lowest terms by extracting and canceling out 3.

$$\sqrt{\frac{448}{3}}$$

Rewrite the square root of the division \sqrt{\frac{448}{3}} as the division of square roots \frac{\sqrt{448}}{\sqrt{3}}.

$$\frac{\sqrt{448}}{\sqrt{3}}$$

Factor 448=8^{2}\times 7. Rewrite the square root of the product \sqrt{8^{2}\times 7} as the product of square roots \sqrt{8^{2}}\sqrt{7}. Take the square root of 8^{2}.

$$\frac{8\sqrt{7}}{\sqrt{3}}$$

Rationalize the denominator of \frac{8\sqrt{7}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{8\sqrt{7}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{8\sqrt{7}\sqrt{3}}{3}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{8\sqrt{21}}{3}$$