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 $2^{2}$.
$$\frac{3\sqrt{2}}{2\sqrt{2}+\sqrt{5}}$$
Rationalize the denominator of $\frac{3\sqrt{2}}{2\sqrt{2}+\sqrt{5}}$ by multiplying numerator and denominator by $2\sqrt{2}-\sqrt{5}$.
Consider $\left(2\sqrt{2}+\sqrt{5}\right)\left(2\sqrt{2}-\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
