Answer
The inequality is valid for values between -8 and 2 (not including them) i.e. $(-8,2)$
Work Step by Step
First, we are going to set the right side to zero and factor to find the x-intercepts:
$x^2+6x-16=0$
$(x+8)(x-2)=0$
$x_1=-8$
$x_2=2$
These are the critical points. We are going to take three values: one less than -8, one between -8 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 -8:
$(-10)^2+6(-10)-16<0$
$100-60-16<0$
$24<0 \rightarrow \text{ FALSE}$
Second test with a value between -8 and 2:
$(0)^2+6(0)-16<0$
$0-0-16<0$
$-16<0 \rightarrow \text{ TRUE}$
Third test with a value more than 2:
$(3)^2+6(3)-16<0$
$9+18-16<0$
$11<0 \rightarrow \text{ FALSE}$
These tests show that the inequality $x^2+6x-16<0$ is valid for values between -8 and 2 (not including them) i.e. $(-8,2)$
