\(x<-3 \lor x > 5\)

We have the inequality \[x^2-15 > 2 x\] but first we solve the following equation: \(x^2-15 = 2 x \), that is \( x^2-2 x-15 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{2\pm \sqrt{(2)^2-4 \cdot 1 \cdot -15}}{2}\\ \\ &=\frac{2\pm \sqrt{64}}{2}\\ \\ &=\frac{2\pm 8}{2}\end{aligned}\] So \[x=-3\quad \text{or}\quad x=5\] Now we explore where the inequality is true.
First we take a value \(x<-3\), say \(x=-5\). The value of the left-hand side of the inequality is then \[(-5)^2-15=10\] The value of the right-hand side is \[2 \cdot -5=-10\] So we have found for \(x<-3\) that \(x^2-15 > 2 x\).
Next we choose a value \(-3<x<5\), say \(x=-2\). The value of the left-hand side of the inequality is then \[(-2)^2-15=-11\] The value of the right-handside is \[2\cdot -2=-4\] So we have found for \(-3<x<5\) that \(x^2-15 < 2 x\).
Finally we choose a value \(x>5\), say \(x=6\). The value of the left-hand side of the inequality is then \[(6)^2-15=21\] The value of the right-hand side is \[2 \cdot 6=12\] So we have found for \(x>5\) that \(x^2-15>2 x\).

So we can conclude that \[x^2-15 > 2 x\] when \(x<-3\) or \(x>5\).