Answer
The inequality is valid for values less than -4 and more than 3 (not including them) i.e. $(-\infty,-4)\cap (3,\infty)$
Work Step by Step
First, we are going to set the right side to zero and factor to find the x-intercepts:
$x^2+x-12=0$
$(x+4)(x-3)=0$
$x_1=-4$
$x_2=3$
These are the critical points. We are going to take three values: one less than -4, one between -4 and 3, and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than -4:
$(-5)^2+(-5)>12$
$25-4>12$
$21>12 \rightarrow \text{ TRUE}$
Second test with a value between -4 and 3:
$(0)^2+0>12$
$0+0>12$
$0>12 \rightarrow \text{ FALSE}$
Third test with a value more than 3:
$(4)^2+4>12$
$16+4>12$
$20>12 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^2+x>12$ is valid for values less than -4 and more than 3 (not including them) i.e. $(-\infty,-4)\cap (3,\infty)$