To solve the equation

\[
\frac{2}{x-2} + \frac{3}{x-3} = \frac{3x-20}{(x-2)(x-3)},
\]

we will first find a common denominator for the fractions on the left-hand side. The common denominator is \((x-2)(x-3)\). Rewriting each fraction with this common denominator, we get:

\[
\frac{2(x-3)}{(x-2)(x-3)} + \frac{3(x-2)}{(x-2)(x-3)} = \frac{3x-20}{(x-2)(x-3)}.
\]

Now, combine the fractions on the left-hand side:

\[
\frac{2(x-3) + 3(x-2)}{(x-2)(x-3)} = \frac{3x-20}{(x-2)(x-3)}.
\]

Simplify the numerator on the left-hand side:

\[
2(x-3) + 3(x-2) = 2x - 6 + 3x - 6 = 5x - 12.
\]

So the equation becomes:

\[
\frac{5x - 12}{(x-2)(x-3)} = \frac{3x - 20}{(x-2)(x-3)}.
\]

Since the denominators are the same, we can equate the numerators:

\[
5x - 12 = 3x - 20.
\]

Solve for \(x\) by isolating \(x\) on one side of the equation:

\[
5x - 3x = -20 + 12,
\]

\[
2x = -8,
\]

\[
x = -4.
\]

Now, we need to check if \(x = -4\) is a valid solution by ensuring it does not make any denominator zero. The denominators are \(x-2\) and \(x-3\):

\[
x - 2 \neq 0 \quad \text{and} \quad x - 3 \neq 0,
\]

\[
-4 - 2 \neq 0 \quad \text{and} \quad -4 - 3 \neq 0,
\]

\[
-6 \neq 0 \quad \text{and} \quad -7 \neq 0.
\]

Since neither denominator is zero, \(x = -4\) is a valid solution. Therefore, the solution to the equation is:

\[
x = -4.
\]