Solve (sqrt{3/8}-2sqrt{3})*sqrt{6}+sqrt{72}=

$$( \sqrt { \frac { 3 } { 8 } } - 2 \sqrt { 3 } ) \times \sqrt { 6 } + \sqrt { 72 } =$$

Evaluate

$$\frac{3}{2}=1.5$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{3}{8}} as the division of square roots \frac{\sqrt{3}}{\sqrt{8}}.

$$\left(\frac{\sqrt{3}}{\sqrt{8}}-2\sqrt{3}\right)\sqrt{6}+\sqrt{72}$$

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

$$\left(\frac{\sqrt{3}}{2\sqrt{2}}-2\sqrt{3}\right)\sqrt{6}+\sqrt{72}$$

Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(\frac{\sqrt{3}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-2\sqrt{3}\right)\sqrt{6}+\sqrt{72}$$

The square of \sqrt{2} is 2.

$$\left(\frac{\sqrt{3}\sqrt{2}}{2\times 2}-2\sqrt{3}\right)\sqrt{6}+\sqrt{72}$$

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

$$\left(\frac{\sqrt{6}}{2\times 2}-2\sqrt{3}\right)\sqrt{6}+\sqrt{72}$$

Multiply 2 and 2 to get 4.

$$\left(\frac{\sqrt{6}}{4}-2\sqrt{3}\right)\sqrt{6}+\sqrt{72}$$

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

$$\left(\frac{\sqrt{6}}{4}+\frac{4\left(-2\right)\sqrt{3}}{4}\right)\sqrt{6}+\sqrt{72}$$

Since \frac{\sqrt{6}}{4} and \frac{4\left(-2\right)\sqrt{3}}{4} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{6}+4\left(-2\right)\sqrt{3}}{4}\sqrt{6}+\sqrt{72}$$

Do the multiplications in \sqrt{6}+4\left(-2\right)\sqrt{3}.

$$\frac{\sqrt{6}-8\sqrt{3}}{4}\sqrt{6}+\sqrt{72}$$

Express \frac{\sqrt{6}-8\sqrt{3}}{4}\sqrt{6} as a single fraction.

$$\frac{\left(\sqrt{6}-8\sqrt{3}\right)\sqrt{6}}{4}+\sqrt{72}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 6\sqrt{2} times \frac{4}{4}.

$$\frac{\left(\sqrt{6}-8\sqrt{3}\right)\sqrt{6}}{4}+\frac{4\times 6\sqrt{2}}{4}$$

Since \frac{\left(\sqrt{6}-8\sqrt{3}\right)\sqrt{6}}{4} and \frac{4\times 6\sqrt{2}}{4} have the same denominator, add them by adding their numerators.

$$\frac{\left(\sqrt{6}-8\sqrt{3}\right)\sqrt{6}+4\times 6\sqrt{2}}{4}$$

Do the multiplications in \left(\sqrt{6}-8\sqrt{3}\right)\sqrt{6}+4\times 6\sqrt{2}.

$$\frac{6-24\sqrt{2}+24\sqrt{2}}{4}$$

Do the calculations in 6-24\sqrt{2}+24\sqrt{2}.

$$\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$\frac{3}{2}$$

Factor

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$