Answer
$\left\{-6,-5\right\}$
Work Step by Step
Let $z=
(x+3)
$. Then the given equation, $
(x+3)^2+5(x+3)+6=0
,$ is equivalent to
\begin{align*}
z^2+5z+6=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+2)(z+3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+2=0 & z+3=0
\\
z=-2 & z=-3
.\end{array}
Since $z=
(x+3)
,$ by back substitution, then
\begin{array}{l|r}
x+3=-2 & x+3=-3
\\
x=-2-3 & x=-3-3
\\
x=-5 & x=-6
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=-5: & \text{If }x=-6
\\\\
(-5+3)^2+5(-5+3)+6\overset{?}=0 &
(-6+3)^2+5(-6+3)+6\overset{?}=0
\\
(-2)^2+5(-2)+6\overset{?}=0 &
(-3)^2+5(-3)+6\overset{?}=0
\\
4-10+6\overset{?}=0 &
9-15+6\overset{?}=0
\\
0\overset{\checkmark}=0 &
0\overset{\checkmark}=0
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
(x+3)^2+5(x+3)+6=0
$ is $\left\{-6,-5\right\}$.
