Answer
The solutions are $x=-14$ or $x=0$.
Work Step by Step
$g(x)=x^{2}+14x+49$
$ g(x)=49\qquad$...substitute $49$ for $g(x)$.
$49=x^{2}+14x+49$
Apply Perfect square formula:
$(x+a)^{2}=x^{2}+2ax+a^{2}$,
$49=(x+7)^{2}$
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+7=\pm\sqrt{49}\qquad$...add $-7$ to each side.
$ x+7-7=\pm\sqrt{49}-7\qquad$...simplify.
$x=-7\pm 7$
$x=0$ or $x=-14$
