Answer
$\left\{ \dfrac{5-\sqrt{33}}{2},\dfrac{5+\sqrt{33}}{2} \right\}$
Work Step by Step
Factoring the given expression and then multiplying both sides of the given equation, $
\dfrac{11}{2x^2+x-15}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}
,$ by the $LCD=
(2x-5)(x+3)
$, then,
\begin{array}{l}\require{cancel}
\dfrac{11}{(2x-5)(x+3)}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}
\\\\
(2x-5)(x+3)\left( \dfrac{11}{(2x-5)(x+3)} \right)=\left( \dfrac{5}{2x-5}-\dfrac{x}{x+3} \right)(2x-5)(x+3)
\\\\
1(11)=5(x+3)-x(2x-5)
\\\\
11=5x+15-2x^2+5x
\\\\
2x^2+(-5x-5x)+(11-15)=0
\\\\
2x^2-10x-4=0
\\\\
\dfrac{2x^2-10x-4}{2}=\dfrac{0}{2}
\\\\
x^2-5x-2=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2-5x-2=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(-2)}}{2(1)}
\\\\=
\dfrac{5\pm\sqrt{25+8}}{2}
\\\\=
\dfrac{5\pm\sqrt{33}}{2}
.\end{array}
Upon checking, both solutions satisfy the original equation. Hence, the solutions are $
\left\{ \dfrac{5-\sqrt{33}}{2},\dfrac{5+\sqrt{33}}{2} \right\}
.$
