Answer
The solutions are $x=-12$ and $x=-1$
Work Step by Step
$(x+3)^{2}+7(x+3)-18=0$
Let $u$ be equal to $x+3$
If $u=x+3$, then $u^{2}=(x+3)^{2}$
Rewrite the original equation using the new variable $u$:
$u^{2}+7u-18=0$
Solve by factoring:
$(u+9)(u-2)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u+9=0$
$u=-9$
$u-2=0$
$u=2$
Substitute $u$ back to $x+3$ and solve for $x$:
$x+3=-9$
$x=-9-3$
$x=-12$
$x+3=2$
$x=2-3$
$x=-1$
The solutions are $x=-12$ and $x=-1$
