#### Answer

The solution set is $\left\{-6, 1\right\}$.
This means that both $-6$ 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 $b=5$ and $c=-6$.
Note that $-6=6(-1)$ and $5 = 6+(-1)$.
This means that $d=6$ and $e=-1$
Thus, the factored form of the trinomial is: $(m+6)[m+(-1)]= (m+6)(m-1)$
The given equation can be written as:
$(m+6)(m-1)=0$
Use the Zero-Product Property by equating each factor to zero.
Then, solve each equation to obtain:
$\begin{array}{ccc}
&m+6 = 0 &\text{ or } &m-1=0
\\&m=-6 &\text{ or } &m=1
\end{array}$
Thus, the solution set is $\left\{-6, 1\right\}$.
This means that both $-6$ and $1$ satisfy the equation.