Answer
The solution set is $\left\{-1\right\}$.
Work Step by Step
Raise both sides of the equation to the 5th power to obtain:
$x^2+2x = (-1)^5
\\x^2+2x=-1$
Add 1 on both sides of the equation to obtain:
$x^2+2x+1=0$
With $a=1, b=2, c=1$, use the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ to obtain:
$x=\dfrac{-2 \pm \sqrt{2^2-4(1)(1)}}{2(1)}
\\x=\dfrac{-2\pm\sqrt{4-4}}{2}
\\x=\dfrac{-2\pm0}{2}
\\x=-1$
Thus, the solution set is $\left\{-1\right\}$.
