To solve the equation

\[
\frac{3}{x+3} - \frac{1}{x-2} = \frac{5}{2x+6}
\]

we can start by simplifying the right-hand side. Notice that \(2x + 6\) can be factored as \(2(x + 3)\):

\[
\frac{5}{2x+6} = \frac{5}{2(x+3)} = \frac{5}{2} \cdot \frac{1}{x+3} = \frac{5}{2(x+3)}
\]

So the equation becomes:

\[
\frac{3}{x+3} - \frac{1}{x-2} = \frac{5}{2(x+3)}
\]

Next, let's find a common denominator for the fractions on the left-hand side. The common denominator for \(x+3\) and \(x-2\) is \((x+3)(x-2)\). Rewrite each fraction with this common denominator:

\[
\frac{3(x-2)}{(x+3)(x-2)} - \frac{1(x+3)}{(x+3)(x-2)} = \frac{5}{2(x+3)}
\]

Simplify the numerators:

\[
\frac{3x - 6}{(x+3)(x-2)} - \frac{x + 3}{(x+3)(x-2)} = \frac{5}{2(x+3)}
\]

Combine the fractions on the left-hand side:

\[
\frac{3x - 6 - (x + 3)}{(x+3)(x-2)} = \frac{5}{2(x+3)}
\]

Simplify the numerator:

\[
\frac{3x - 6 - x - 3}{(x+3)(x-2)} = \frac{5}{2(x+3)}
\]

\[
\frac{2x - 9}{(x+3)(x-2)} = \frac{5}{2(x+3)}
\]

Now, cross-multiply to solve for \(x\):

\[
2(2x - 9) = 5(x - 2)
\]

Distribute the constants:

\[
4x - 18 = 5x - 10
\]

Rearrange the equation to isolate \(x\):

\[
4x - 5x = -10 + 18
\]

\[
-x = 8
\]

\[
x = -8
\]

Finally, check the solution by substituting \(x = -8\) back into the original equation to ensure it does not make any denominator zero and satisfies the equation:

\[
\frac{3}{-8+3} - \frac{1}{-8-2} = \frac{5}{2(-8)+6}
\]

\[
\frac{3}{-5} - \frac{1}{-10} = \frac{5}{-16+6}
\]

\[
-\frac{3}{5} + \frac{1}{10} = \frac{5}{-10}
\]

Find a common denominator for the left-hand side:

\[
-\frac{6}{10} + \frac{1}{10} = -\frac{5}{10} = -\frac{1}{2}
\]

And the right-hand side:

\[
\frac{5}{-10} = -\frac{1}{2}
\]

Both sides are equal, so the solution \(x = -8\) is correct.