Answer
the solution set is\[\left\{ 6 \right\}\]
Work Step by Step
Step 1:Shift all nonzero terms to left side and obtain zero on the other side.
For this, add 36 both sides as follows:
\[{{x}^{2}}-12x+36=-36+36\]
This implies that,
\[{{x}^{2}}-12x+36=0\]
So, after shifting all nonzero terms to the left side, the above equation becomes: \[{{x}^{2}}-12x+36=0\]
Step 2:Find the factor of the above equation:
Consider the equation\[{{x}^{2}}-12x+36=0\].
Factorize it as follows:
\[\begin{align}
& {{x}^{2}}-12x+36=0 \\
& {{x}^{2}}-6x-6x+36=0 \\
& x\left( x-6 \right)-6\left( x-6 \right)=0 \\
& \left( x-6 \right)\left( x-6 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( x-6 \right)\left( x-6 \right)=0\].
By the zero product principal, \[\left( x-6 \right)=0\].
That is,\[x=6\]repeated twice.
Step5: Check the solution in the original equation:
Check for\[x=6\]. So consider,
\[\begin{align}
& {{x}^{2}}-12x+36=0 \\
& {{\left( 6 \right)}^{2}}-12\left( 6 \right)+36=0 \\
& 36-72+36=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ 6 \right\}\].
