Answer
The solutions are $x=-6+\sqrt{15}$ or $x=-6-\sqrt{15}$.
Work Step by Step
$f(t)=(t+6)^{2}$
$ f(t)=15\qquad$...substitute $15$ for $f(t)$.
$(t+6)^{2}=15$
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+6=\pm\sqrt{15}\qquad$...add $-6$ to each side.
$ x+6-6=\pm\sqrt{15}-6\qquad$...simplify.
$x=-6\pm\sqrt{15}$
$x=-6+\sqrt{15}$ or $x=-6-\sqrt{15}$