Answer
The inequality is valid on values less than -5 and more than 2 (not including them) i.e. $(-\infty,-5)\cap (2,\infty)$
Work Step by Step
First, we are going to set the right side to zero and factor to find the x-intercepts:
$x^2+3x-10$
$(x+5)(x-2)$
$x_1=-5$
$x_2=2$
These are the critical points. We are going to take three values: one less than -5, one between -5 and 2, and one more than 2 to test in the original equation and check if the inequality is true or not:
First test with a value less than -5:
$(-6)^2+3(-6)-10>0$
$36-18-10>0$
$8>0 \rightarrow \text{ TRUE}$
Second test with a value between -5 and 2:
$(0)^2+3(0)-10>0$
$9-9-10>0$
$-10>0 \rightarrow \text{ FALSE}$
Third test with a value more than 2:
$(3)^2+3(3)-10>0$
$9+9-10>0$
$8>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^2+3x-10>0$ is valid on values less than -5 and more than 2 (not including them) i.e. $(-\infty,-5)\cap (2,\infty)$