Answer
The inequality is valid on values less than -8 and more than 0 (not including them) i.e. $(-\infty,-8)\cap (0,\infty)$
Work Step by Step
First, we are going to set the right side to zero and factor to find the x-intercepts:
$x^2+8x$
$x(x+8)$
$x_1=-8$
$x_2=0$
These are the critical points. We are going to take three values: one less than -8, one between -8 and 0, and one more than 0 to test in the original equation and check if the inequality is true or not:
First test with a value less than -8:
$(-10)^2+8(-10)>0$
$100-80>0$
$20>0 \rightarrow \text{ TRUE}$
Second test with a value between -8 and 0:
$(-1)^2+8(-1)>0$
$1-8>0$
$-9>0 \rightarrow \text{ FALSE}$
Third test with a value more than 0:
$3^2+8(3)>0$
$9+24>0$
$33>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^2+8x>0$ is valid on values less than -8 and more than 0 (not including them) i.e. $(-\infty,-8)\cap (0,\infty)$
