Answer
The solutions are $x=-4+\sqrt{13}$ or $x=-4-\sqrt{13}$.
Work Step by Step
$F(t)=(t+4)^{2}$
$ F(t)=13\qquad$...substitute $13$ for $F(t)$.
$(t+4)^{2}=13$
According to the general principle of square roots:
For any real number $k$ and any algebraic expression $x$ :
$\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$.
$ t+4=\pm\sqrt{13}\qquad$...add $-4$ to each side.
$ x+4-4=\pm\sqrt{13}-4\qquad$...simplify.
$x=-4\pm\sqrt{13}$
$x=-4+\sqrt{13}$ or $x=-4-\sqrt{13}$
