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

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 \[x=-5\quad \text{or}\quad x=3\] Now we explore where the inequality is true.
First we take a value \(x<-5\), say \(x=-7\). The value of the left-hand side of the inequality is then \[(-7)^2-15=34\] The value of the right-hand side is \[-2 \cdot -7=14\] So we have found for \(x<-5\) that \(x^2-15 > -2 x\).
Next we choose a value \(-5<x<3\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-15=1\] The value of the right-handside is \[-2\cdot -4=8\] So we have found for \(-5<x<3\) that \(x^2-15 < -2 x\).
Finally we choose a value \(x>3\), say \(x=4\). The value of the left-hand side of the inequality is then \[(4)^2-15=1\] The value of the right-hand side is \[-2 \cdot 4=-8\] So we have found for \(x>3\) that \(x^2-15>-2 x\).

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