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