Answer
The solutions are $x=1$ or $x=9$
Work Step by Step
$f(x)=(x-5)^{2}$
$f(x)=16\qquad$...substitute $16$ for $f(x)$
$(x-5)^{2}=16$
According to the general principle of square roots:
For any real number $k$ and any algebraic expression $x$ :
$\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$.
$ x-5=\pm\sqrt{16}\qquad$...add $5$ to each side.
$ x-5+5=\pm\sqrt{16}+5\qquad$...simplify.
$x=5\pm 4$
$x=5+4$ or $x=5-4$
$x=9$ or $x=1$
