Answer
The solutions are $7$ or $-1$.
Work Step by Step
$(x-3)^{2}=16$
According to the general case of the 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-3=\pm\sqrt{16}\qquad$...add $3$ to each side.
$ x-3+3=\pm\sqrt{16}+3\qquad$...simplify.
$x=3\pm 4$
$x=3+4$ or $x=3-4$
$x=7$ or $x=-1$
