Solve sqrt{48}divsqrt{3}+2sqrt{1/5}*sqrt{30}-(2sqrt{2}+sqrt{3})^2

$$\sqrt { 48 } \div \sqrt { 3 } + 2 \sqrt { \frac { 1 } { 5 } } \times \sqrt { 30 } - ( 2 \sqrt { 2 } + \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$-2\sqrt{6}-7\approx -11.898979486$$

Short Solution Steps

Rewrite the division of square roots \frac{\sqrt{48}}{\sqrt{3}} as the square root of the division \sqrt{\frac{48}{3}} and perform the division.

$$\sqrt{16}+2\sqrt{\frac{1}{5}}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Calculate the square root of 16 and get 4.

$$4+2\sqrt{\frac{1}{5}}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{1}{5}} as the division of square roots \frac{\sqrt{1}}{\sqrt{5}}.

$$4+2\times \frac{\sqrt{1}}{\sqrt{5}}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Calculate the square root of 1 and get 1.

$$4+2\times \frac{1}{\sqrt{5}}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Rationalize the denominator of \frac{1}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$4+2\times \frac{\sqrt{5}}{\left(\sqrt{5}\right)^{2}}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4+2\times \frac{\sqrt{5}}{5}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Express 2\times \frac{\sqrt{5}}{5} as a single fraction.

$$4+\frac{2\sqrt{5}}{5}\sqrt{30}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Express \frac{2\sqrt{5}}{5}\sqrt{30} as a single fraction.

$$4+\frac{2\sqrt{5}\sqrt{30}}{5}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{5}{5}.

$$\frac{4\times 5}{5}+\frac{2\sqrt{5}\sqrt{30}}{5}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Since \frac{4\times 5}{5} and \frac{2\sqrt{5}\sqrt{30}}{5} have the same denominator, add them by adding their numerators.

$$\frac{4\times 5+2\sqrt{5}\sqrt{30}}{5}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Do the multiplications in 4\times 5+2\sqrt{5}\sqrt{30}.

$$\frac{20+10\sqrt{6}}{5}-\left(2\sqrt{2}+\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{2}+\sqrt{3}\right)^{2}.

$$\frac{20+10\sqrt{6}}{5}-\left(4\left(\sqrt{2}\right)^{2}+4\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$\frac{20+10\sqrt{6}}{5}-\left(4\times 2+4\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

Multiply 4 and 2 to get 8.

$$\frac{20+10\sqrt{6}}{5}-\left(8+4\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{20+10\sqrt{6}}{5}-\left(8+4\sqrt{6}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$\frac{20+10\sqrt{6}}{5}-\left(8+4\sqrt{6}+3\right)$$

Add 8 and 3 to get 11.

$$\frac{20+10\sqrt{6}}{5}-\left(11+4\sqrt{6}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 11+4\sqrt{6} times \frac{5}{5}.

$$\frac{20+10\sqrt{6}}{5}-\frac{5\left(11+4\sqrt{6}\right)}{5}$$

Since \frac{20+10\sqrt{6}}{5} and \frac{5\left(11+4\sqrt{6}\right)}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{20+10\sqrt{6}-5\left(11+4\sqrt{6}\right)}{5}$$

Do the multiplications in 20+10\sqrt{6}-5\left(11+4\sqrt{6}\right).

$$\frac{20+10\sqrt{6}-55-20\sqrt{6}}{5}$$

Do the calculations in 20+10\sqrt{6}-55-20\sqrt{6}.

$$\frac{-35-10\sqrt{6}}{5}$$

Divide each term of -35-10\sqrt{6} by 5 to get -7-2\sqrt{6}.

$$-7-2\sqrt{6}$$