Question 1066741
GIVEN  {{{highlight(x^2+x-12=2x)}}}
(and decide if you want the variable X or x.  They are not the same.)


{{{x^2+x-12-2x=0}}}
{{{x^2-x-12=0}}}
{{{(x+3)(x-4)=0}}}

Your found roots of -3 and +4 are correct.


{{{f(x)=x^2-x-12=(x+3)(x-4)}}}  is the function.  It has the roots which you found.  You can find the value in the middle of the roots and use this to find the <s>maximum</s> minimum or vertex of the related function.


Better Understanding:
The functions requested are from your original given equation.

{{{g(x)=x^2+x-12}}} and {{{h(x)=2x}}}


{{{graph(300,300,-8,8,-8,8,x^2+x-12,2x)}}}
Notice the x-values where the related functions intersect.