Answer
The solutions are $x=-7$ or $x=-1$.
Work Step by Step
$F(x)=x^{2}+8x+16$
$ F(x)=9\qquad$...substitute $9$ for $F(x)$.
$9=x^{2}+8x+16$
Apply Perfect square formula:
$(x+a)^{2}=x^{2}+2ax+a^{2}$,
$9=(x+4)^{2}$
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+4=\pm\sqrt{9}\qquad$...add $-4$ to each side.
$ x+4-4=\pm\sqrt{9}-4\qquad$...simplify.
$x=-4\pm 3$
$x=-4+3$ or $x=-4-3$
$x=-1$ or $x=-7$
