Answer
The solutions are $x=-5$ and $x=-3$
Work Step by Step
$x^{2}+8x=-15$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=8$:
$x^{2}+8x+\Big(\dfrac{8}{2}\Big)^{2}=-15+\Big(\dfrac{8}{2}\Big)^{2}$
$x^{2}+8x+16=-15+16$
$x^{2}+8x+16=1$
Factor the left side of the equation, which is a perfect square trinomial:
$(x+4)^{2}=1$
Take the square root of both sides:
$\sqrt{(x+4)^{2}}=\pm\sqrt{1}$
$x+4=\pm1$
Solve for $x$:
$x=\pm1-4$
$x=-1-4=-5$
$x=1-4=-3$
The solutions are $x=-5$ and $x=-3$
