Answer
$x=\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$
Work Step by Step
$\dfrac{11}{2x^{2}+x-15}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}$
Factor the denominator of the fraction on the left side of the equation:
$\dfrac{11}{(2x-5)(x+3)}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}$
Multiply the whole equation by $(2x-5)(x+3)$:
$(2x-5)(x+3)\Big[\dfrac{11}{(2x-5)(x+3)}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}\Big]$
$11=5(x+3)-x(2x-5)$
$11=5x+15-2x^{2}+5x$
Take all terms to the left side of the equation and simplify it:
$11-5x-15+2x^{2}-5x=0$
$2x^{2}-10x-4=0$
Take out common factor $2$ from this equation and simplify it further by taking the common factor to divide the right side of the equation:
$2(x^{2}-5x-2)=0$
$x^{2}-5x-2=\dfrac{0}{2}$
$x^{2}-5x-2=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-5$ and $c=-2$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(-2)}}{2(1)}=\dfrac{5\pm\sqrt{25+8}}{2}=...$
$...=\dfrac{5\pm\sqrt{33}}{2}=\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$
The original equation is not undefined for neither of the values of $x$ found. Our final answer is $x=\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$
