Answer
$-2 \lt x \lt 4$
The solution set is $~~(-2,4)$
Work Step by Step
$x^2 \lt 2x+8$
$x^2 -2x-8 \lt 0$
$(x-4)(x+2) \lt 0$
When $~~x=-2~~$ or $~~x=4,~~$ then $~~(x-4)(x+2) = 0$
When $x \lt -2,~~$ then $~~(x-4)(x+2) \gt 0$
When $-2 \lt x \lt 4,~~$ then $~~(x-4)(x+2) \lt 0$
When $x \gt 4,~~$ then $~~(x-4)(x+2) \gt 0$
Solution:
$-2 \lt x \lt 4$
The solution set is $~~(-2,4)$
