\(x<-4 \lor x > 6\)

We have the inequality \[x^2-24 > 2 x\] but first we solve the following equation: \(x^2-24 = 2 x \), that is \( x^2-2 x-24 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{2\pm \sqrt{(2)^2-4 \cdot 1 \cdot -24}}{2}\\ \\ &=\frac{2\pm \sqrt{100}}{2}\\ \\ &=\frac{2\pm 10}{2}\end{aligned}\] So \[x=-4\quad \text{or}\quad x=6\] 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 \[(-6)^2-24=12\] The value of the right-hand side is \[2 \cdot -6=-12\] So we have found for \(x<-4\) that \(x^2-24 > 2 x\).
Next we choose a value \(-4<x<6\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-24=-15\] The value of the right-handside is \[2\cdot -3=-6\] So we have found for \(-4<x<6\) that \(x^2-24 < 2 x\).
Finally we choose a value \(x>6\), say \(x=7\). The value of the left-hand side of the inequality is then \[(7)^2-24=25\] The value of the right-hand side is \[2 \cdot 7=14\] So we have found for \(x>6\) that \(x^2-24>2 x\).

So we can conclude that \[x^2-24 > 2 x\] when \(x<-4\) or \(x>6\).