Answer
$\left\{-4,1\right\}$
Work Step by Step
Let $z=
(t+5)
$. Then the given equation, $
(t+5)^2+6=7(t+5)
,$ is equivalent to
\begin{align*}
z^2+6&=7z
\\
z^2-7z+6&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-6)(z-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-6=0 & z-1=0
\\
z=6 & z=1
.\end{array}
Since $z=
(t+5)
,$ by back substitution, then
\begin{array}{l|r}
t+5=6 & t+5=1
\\
t=6-5 & t=1-5
\\
t=1 & t=-4
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }t=1: & \text{If }t=-4
\\\\
(1+5)^2+6\overset{?}=7(1+5) &
(-4+5)^2+6\overset{?}=7(-4+5)
\\
6^2+6\overset{?}=7(6) &
(1)^2+6\overset{?}=7(1)
\\
36+6\overset{?}=42 &
1+6\overset{?}=7
\\
42\overset{\checkmark}=42 &
7\overset{\checkmark}=7
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
(t+5)^2+6=7(t+5)
$ is $\left\{-4,1\right\}$.
