Answer
The required solution set is $\left\{ -4 \right\}$.
Work Step by Step
Consider the equation $\frac{2}{x+3}-\frac{4}{x+5}=\frac{6}{{{x}^{2}}+8x+15}$.
Solve it as:
$\begin{align}
& \frac{2}{x+3}-\frac{4}{x+5}=\frac{6}{{{x}^{2}}+8x+15} \\
& \frac{2}{x+3}-\frac{4}{x+5}=\frac{6}{\left( x+3 \right)\left( x+5 \right)} \\
& \frac{2\left( x+5 \right)-4\left( x+3 \right)}{\left( x+3 \right)\left( x+5 \right)}=\frac{6}{\left( x+3 \right)\left( x+5 \right)} \\
& 2\left( x+5 \right)-4\left( x+3 \right)=6
\end{align}$
This implies,
$\begin{align}
& 2\left( x+5 \right)-4\left( x+3 \right)=6 \\
& 2x+10-4x-12=6 \\
& -2x-2=6 \\
& -2x=8
\end{align}$
And
$\begin{align}
& x=-\frac{8}{2} \\
& =-4
\end{align}$
Therefore, the solution set is $\left\{ -4 \right\}$.
