Answer
$\left\{-1,8\right\}$
Work Step by Step
Let $z=
(x-4)
$. Then the given equation, $
(x-4)^2+(x-4)-20=0
,$ is equivalent to
\begin{align*}
z^2+z-20&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+5)(z-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+5=0 & z-4=0
\\
z=-5 & z=4
.\end{array}
Since $z=
(x-4)
,$ by back substitution, then
\begin{array}{l|r}
x-4=-5 & x-4=4
\\
x=-5+4 & x=4+4
\\
x=-1 & x=8
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=-1: & \text{If }x=8
\\\\
(-1-4)^2+(-1-4)-20\overset{?}=0 &
(8-4)^2+(8-4)-20\overset{?}=0
\\
(-5)^2+(-5)-20\overset{?}=0 &
(4)^2+(4)-20\overset{?}=0
\\
25-5-20\overset{?}=0 &
16+4-20\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-4)^2+(x-4)-20=0
$ is $\left\{-1,8\right\}$.
