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

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 \[x=-2\quad \text{or}\quad x=4\] Now we explore where the inequality is true.
First we take a value \(x<-2\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-8=8\] The value of the right-hand side is \[2 \cdot -4=-8\] So we have found for \(x<-2\) that \(x^2-8 > 2 x\).
Next we choose a value \(-2<x<4\), say \(x=-1\). The value of the left-hand side of the inequality is then \[(-1)^2-8=-7\] The value of the right-handside is \[2\cdot -1=-2\] So we have found for \(-2<x<4\) that \(x^2-8 < 2 x\).
Finally we choose a value \(x>4\), say \(x=5\). The value of the left-hand side of the inequality is then \[(5)^2-8=17\] The value of the right-hand side is \[2 \cdot 5=10\] So we have found for \(x>4\) that \(x^2-8>2 x\).

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