Answer
$\dfrac{8\sqrt 5-19 }{41}$
Work Step by Step
Rationalize the denominator by multiplying $2-3\sqrt5$ to both the numerator and the denominator:
$\dfrac{2-\sqrt 5}{2+3\sqrt 5}\\
=\dfrac{2-\sqrt 5}{2+3\sqrt 5} \cdot \dfrac{2-3\sqrt 5}{2-3\sqrt 5}$
Use FOIL method in the numerator and special formula $(a+b)(a-b)=a^2-b^2$ in the denominator.
$=\dfrac{2\cdot 2 -2\cdot 3\sqrt 5-\sqrt 5\cdot 2+\sqrt 5 \cdot 3\sqrt 5}{2^2-(3\sqrt 5)^2}$
$=\dfrac{4 -6\sqrt 5-2\sqrt 5+3(\sqrt 5)^2}{4-9(\sqrt 5)^2}$
Use $(\sqrt{a})^2=a, a>0$, then simplify to obtain:
$=\dfrac{4 -6\sqrt 5-2\sqrt 5+3(5) }{4-9(5)}$
$=\dfrac{4 -6\sqrt 5-2\sqrt 5+15}{4-45}$
$=\dfrac{19 -6\sqrt 5-2\sqrt 5}{-41}$
Add like terms.
$=\dfrac{19 -8\sqrt 5 }{-41}$
Multiply both the numerator and the denominator by $-1$:
$=\dfrac{-1(19 -8\sqrt 5) }{-1(-41)}$
$=\dfrac{-19 +8\sqrt 5) }{41}$
$=\dfrac{8\sqrt 5-19 }{41}$
Hence, the correct answer is $\frac{8\sqrt 5-19 }{41}$.
