Answer
$x=-\dfrac{5}{2}\pm\dfrac{\sqrt{17}}{2}$
Work Step by Step
$x^{2}+5x=-2$
Take the $-2$ to the left side of the equation:
$x^{2}+5x+2=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=5$ and $c=2$
Substitute:
$x=\dfrac{-5\pm\sqrt{5^{2}-4(1)(2)}}{2(1)}=\dfrac{-5\pm\sqrt{25-8}}{2}=...$
$...=\dfrac{-5\pm\sqrt{17}}{2}=-\dfrac{5}{2}\pm\dfrac{\sqrt{17}}{2}$
