Rationalize the denominator of the given expression.

To rationalize $\frac{\sqrt{3}}{2-\sqrt{5}}$, multiply the numerator and denominator by the conjugate of the denominator, which is $2+\sqrt{5}$.

$$\frac{\sqrt{3}}{2-\sqrt{5}} = \frac{\sqrt{3}}{2-\sqrt{5}} \times \frac{2+\sqrt{5}}{2+\sqrt{5}}$$

Multiply the numerators:

$$\sqrt{3}(2+\sqrt{5}) = 2\sqrt{3} + \sqrt{3}\sqrt{5} = 2\sqrt{3} + \sqrt{15}$$

Multiply the denominators using the difference of squares formula $(x-y)(x+y) = x^2 - y^2$:

$$(2-\sqrt{5})(2+\sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1$$

We have $-2\sqrt{3} - \sqrt{15}$. This can be written as $-2\sqrt{3} + (-1)\sqrt{15}$.
Comparing this to $a\sqrt{b} + c\sqrt{d}$:
$a = -2$
$b = 3$
$c = -1$
$d = 15$

All values $a, b, c, d$ are rational numbers.

The values are:
$a = \boxed{-2}$
$b = \boxed{3}$
$c = \boxed{-1}$
$d = \boxed{15}$