#### Answer

The solution set is $\left\{-12, 1\right\}$.
This means that both $-12$ and $1$ satisfy the equation.

#### Work Step by Step

RECALL:
A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$.
The trinomial's factored form will be:
$x^2+bx+c=(x+d)(x+e)$
The trinomial in the given equation has $b=11$ and $c=-12$.
Note that $-12=12(-1)$ and $11 = 12+(-1)$.
This means that $d=12$ and $e=-1$
Thus, the factored form of the trinomial is: $(p+12)[p+(-1)]= (p+12)(p-1)$
The given equation can be written as:
$(p+12)(p-1)=0$
Use the Zero-Product Property by equating each factor to zero.
Then, solve each equation to obtain:
$\begin{array}{ccc}
&p+12 = 0 &\text{ or } &p-1=0
\\&p=-12 &\text{ or } &p=1
\end{array}$
Thus, the solution set is $\left\{-12, 1\right\}$.
This means that both $-12$ and $1$ satisfy the equation.