Answer
The solutions are $t=-12$ and $t=-3$.
Work Step by Step
$t^{2}+15t=-36\implies t^{2}+15t+36=0$
To factor the polynomial $t^{2}+bt+c$ to $(t+p)$ and $(t+q)$, we need to have $p+q=b$ and $pq=c$.
In this case, $b=15$ and $c=36$.
$\implies p=12$ and $q=3$.
Then, $t^{2}+15t+36=(t+12)(t+3)$
$t^{2}+15t+36=0\implies (t+12)(t+3)=0$
Using zero-product property, we have
$t+12=0$ or $t+3=0$
$\implies t=-12$ or $t=-3$
The solutions are $t=-12$ and $t=-3$.
