## Intermediate Algebra (6th Edition)

The solutions are $x=-5$ and $x=-3$
$x^{2}+8x=-15$ 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 $x=-3$