Step-by-step explanation:
[tex]x^2+x-12>0\\\\x^2+4x-3x-12>0\\\\x(x+4)-3(x+4)>0\\\\(x+4)(x-3)>0\\\\\text{zeros}\\\\x+4=0\Rightarrow x=-4\\\\x-3=0\Rightarrow x=3[/tex]
Mark the zeros on a number line. Draw a parabola open up that goes through zeros.
We are looking for numerical intervals in which the parabola has positive values (graph above the axis).
(look at the picture)
[tex]x^2+x-12\iff x\in(-\infty,\ -4)\ \cup\ (3,\ \infty)[/tex]
