Answer
The solution set is \[\left\{ -4,8 \right\}\].
Work Step by Step
Step 1: Shift all nonzero terms to left sides and obtain zero on the other side.
For this, subtract 32 from both sides to get \[{{x}^{2}}-4x-32=0\].
Step 2: Find the factor of the above equation:
Consider the equation, \[{{x}^{2}}-4x-32=0\]
Factorize it as follows:
\[\begin{align}
& {{x}^{2}}-4x-32=0 \\
& {{x}^{2}}-8x+4x-32=0 \\
& x\left( x-8 \right)+4\left( x-8 \right)=0 \\
& \left( x+4 \right)\left( x-8 \right)=0
\end{align}\]
Step 3 and 4: Set each factor equal to zero and solve the resulting equation:
From Step 2, \[\left( x+4 \right)\left( x-8 \right)=0\]
By the zero product principal either \[\left( x+4 \right)=0\] or \[\left( x-8 \right)=0\].
Now \[\left( x+4 \right)=0\] implies that \[x=-4\] and \[\left( x-8 \right)=0\] implies that \[x=8\].
Step 5: Check the solution in the original equation:
Check for\[x=-4\].
So consider,
\[\begin{align}
& {{x}^{2}}-4x-32=0 \\
& {{\left( -4 \right)}^{2}}-4\left( -4 \right)-32=0 \\
& 16+16-32=0 \\
& 0=0
\end{align}\]
and, check for\[x=8\].
\[\begin{align}
& {{x}^{2}}-4x-32=0 \\
& {{\left( 8 \right)}^{2}}-4\left( 8 \right)-32=0 \\
& 64-32-32=0 \\
& 0=0
\end{align}\]
