Answer
$x=-7.5 $, $x= 3$
Work Step by Step
Given \begin{equation}
\frac{5}{x+2}+\frac{2 x}{x-7}=\frac{10}{x^2-5 x-14}.
\end{equation} Factor the denominator on the right hand side.
\begin{equation}
\begin{aligned}
x^2-5 x-14&=0 \\
x^2+2x-7x-14&= 0\\
x(x+2)-7(x+2)&= 0\\
(x+2)(x-7)&= 0.
\end{aligned}
\end{equation} Rewrite the solution: \begin{equation}
\begin{aligned}
\frac{5}{x+2}+\frac{2 x}{x-7}&=\frac{10}{x^2-5 x-14}\\
\frac{5}{(x+2)}+\frac{2 x}{(x-7)}&=\frac{10}{(x+2)(x-7)}\\
\left( \frac{5}{(x+2)}+\frac{2 x}{(x-7)} \right)\cdot (x+2)(x-7) & =\frac{10(x+2)(x-7)}{(x+2)(x-7)} \\
2x(x+2)+5(x-7)& =10\\
2x^2+4x+5x-35-10&= 0\\
2x^2+9x-45&= 0.
\end{aligned}
\end{equation} We solve the quadratic equation: \begin{equation}
\begin{aligned}
a & =2, b=9, c=-45 \\
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
x & =\frac{-9 \pm \sqrt{9^2-4 \cdot 2(-45)}}{2 \cdot 2}\\
& =\frac{-9\pm 21}{4} \\
\Longrightarrow x& =\frac{-9 + 21}{4}\\
&=3\\
x& =\frac{-9 -21}{4}\\
&=-7.5\\
\end{aligned}
\end{equation} The solution is $x=-7.5 $ , $x= 3$.
