\(x<-4 \lor x > 2\)

We have the inequality \[x^2-8 > -2 x\] but first we solve the following equation: \(x^2-8 = -2 x \), that is \( x^2+2 x-8 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{-2\pm \sqrt{(-2)^2-4 \cdot 1 \cdot -8}}{2}\\ \\ &=\frac{-2\pm \sqrt{36}}{2}\\ \\ &=\frac{-2\pm 6}{2}\end{aligned}\] So \[x=-4\quad \text{or}\quad x=2\] Now we explore where the inequality is true.
First we take a value \(x<-4\), say \(x=-6\). The value of the left-hand side of the inequality is then \[(-6)^2-8=28\] The value of the right-hand side is \[-2 \cdot -6=12\] So we have found for \(x<-4\) that \(x^2-8 > -2 x\).
Next we choose a value \(-4<x<2\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-8=1\] The value of the right-handside is \[-2\cdot -3=6\] So we have found for \(-4<x<2\) that \(x^2-8 < -2 x\).
Finally we choose a value \(x>2\), say \(x=3\). The value of the left-hand side of the inequality is then \[(3)^2-8=1\] The value of the right-hand side is \[-2 \cdot 3=-6\] So we have found for \(x>2\) that \(x^2-8>-2 x\).

So we can conclude that \[x^2-8 > -2 x\] when \(x<-4\) or \(x>2\).