Answer
$\displaystyle -\frac{19-8\sqrt{5}}{41}=\frac{8\sqrt{5}-19}{41}$
Work Step by Step
We rationalize the denominator:
$\displaystyle \frac{2-\sqrt{5}}{2+3\sqrt{5}}=\frac{(2-\sqrt{5})(2-3\sqrt{5})}{(2+3\sqrt{5})(2-3\sqrt{5})}=\frac{(2-\sqrt{5})(2-3\sqrt{5})}{2^2-(3\sqrt{5})^2}=\frac{(2-\sqrt{5})(2-3\sqrt{5})}{4-9*5}=\frac{2*2-2*3\sqrt{5}-2\sqrt{5}+3\sqrt{5}\sqrt{5}}{-41}=\frac{4-6\sqrt{5}-2\sqrt{5}+3*5}{-41}=\frac{19-8\sqrt{5}}{-41}=\frac{8\sqrt{5}-19}{41}$
