we have the expression
[tex]3a^2-3a-36[/tex]step 1
Factor 3
[tex]3(a^2-a-12)[/tex]step 2
equate to zero
[tex]3(a^2-a-12)=0[/tex]step 3
Solve
[tex](a^2-a-12)=0[/tex][tex]\begin{gathered} a^2-a=12 \\ (a^2-a+\frac{1}{4}-\frac{1}{4})=12 \\ (a^2-a+\frac{1}{4})=12+\frac{1}{4} \\ (a^2-a+\frac{1}{4})=\frac{49}{4} \end{gathered}[/tex]Rewrite as perfect squares
[tex](a-\frac{1}{2})^2=\frac{49}{4}[/tex]take the square root on both sides
[tex]\begin{gathered} a-\frac{1}{2}=\pm\frac{7}{2} \\ a=\frac{1}{2}\pm\frac{7}{2} \end{gathered}[/tex]the values of a are
a=4 and a=-3
therefore
[tex]3(a^2-a-12)=3(a-4)(a+3)[/tex]