Answer
$x=3$ and $x=-5$.
Work Step by Step
Comparing $x^{2}+2x-15=0$ with $ax^{2}+bx+c=0$, we see that $a=1, b=2$ and $c=-15$.
Using the quadratic formula, we have
$x=\frac{-b\pm \sqrt {b^{2}-4ac}}{2a}=\frac{-2\pm\sqrt {2^{2}-4(1)(-15)}}{2(1)}$
$=\frac{-2\pm8}{2}$
So, the solutions are $x=\frac{-2+8}{2}=3$ and $x=\frac{-2-8}{2}=-5$.
