Answer
The solutions are $x=2$ and $x=12$
Work Step by Step
$(x-5)^{2}-4(x-5)-21=0$
Let $u$ be equal to $x-5$
If $u=x-5$, then $u^{2}=(x-5)^{2}$
Rewrite the original equation using the new variable $u$:
$u^{2}-4u-21=0$
Solve by factoring:
$(u-7)(u+3)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-7=0$
$u=7$
$u+3=0$
$u=-3$
Substitute $u$ back to $x-5$ and solve for $x$:
$x-5=7$
$x=7+5$
$x=12$
$x-5=-3$
$x=-3+5$
$x=2$
The solutions are $x=2$ and $x=12$
