Solve m=frac{sqrt{180}+sqrt{20}-11sqrt{5}}{sqrt{5}-sqrt{2}}

$$m = \frac { \sqrt { 180 } + \sqrt { 20 } - 11 \sqrt { 5 } } { \sqrt { 5 } - \sqrt { 2 } }$$

Solve for m

$$m=-\left(\sqrt{10}+5\right)\approx -8.16227766$$

Steps for Solving Linear Equation

Factor 180=6^{2}\times 5. Rewrite the square root of the product \sqrt{6^{2}\times 5} as the product of square roots \sqrt{6^{2}}\sqrt{5}. Take the square root of 6^{2}.

$$m=\frac{6\sqrt{5}+\sqrt{20}-11\sqrt{5}}{\sqrt{5}-\sqrt{2}}$$

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$m=\frac{6\sqrt{5}+2\sqrt{5}-11\sqrt{5}}{\sqrt{5}-\sqrt{2}}$$

Combine 6\sqrt{5} and 2\sqrt{5} to get 8\sqrt{5}.

$$m=\frac{8\sqrt{5}-11\sqrt{5}}{\sqrt{5}-\sqrt{2}}$$

Combine 8\sqrt{5} and -11\sqrt{5} to get -3\sqrt{5}.

$$m=\frac{-3\sqrt{5}}{\sqrt{5}-\sqrt{2}}$$

Rationalize the denominator of \frac{-3\sqrt{5}}{\sqrt{5}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}+\sqrt{2}.

$$m=\frac{-3\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}$$

Consider \left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{5}. Square \sqrt{2}.

$$m=\frac{-3\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{5-2}$$

Subtract 2 from 5 to get 3.

$$m=\frac{-3\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{3}$$

Cancel out 3 and 3.

$$m=-\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)$$

Use the distributive property to multiply -\sqrt{5} by \sqrt{5}+\sqrt{2}.

$$m=-\left(\sqrt{5}\right)^{2}-\sqrt{5}\sqrt{2}$$

The square of \sqrt{5} is 5.

$$m=-5-\sqrt{5}\sqrt{2}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$m=-5-\sqrt{10}$$

Assign m

$$m≔-\left(\sqrt{10}+5\right)$$