Answer :
Given the equation :
[tex]y=x^2-12x+36[/tex]we need to complete the table of x and y
For each value of x, we will substitute with it to find y
[tex]\begin{gathered} x=-6\rightarrow y=(-6)^2-12\cdot-6+36=36+72+36=144 \\ \\ x=5\rightarrow y=(5)^2-12\cdot5+36=25-60+36=1 \\ \\ x=9\rightarrow y=9^2-12\cdot9+36=81-108+36=9 \\ \\ x=4\rightarrow y=4^2-12\cdot4+36=16-48+36=4 \end{gathered}[/tex]We need to find x when y = 36
so,
[tex]\begin{gathered} 36=x^2-12x+36 \\ x^2-12x+36-36=0 \\ x^2-12x=0 \\ x(x-12)=0 \\ x=0\text{ or x -12=0 }\rightarrow x=12 \end{gathered}[/tex]