\(x<-5 \lor x > 4\)

We have the inequality \[x^2-20 > -x\] but first we solve the following equation: \(x^2-20 = -x \), that is \( x^2+x-20 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{-1\pm \sqrt{(-1)^2-4 \cdot 1 \cdot -20}}{2}\\ \\ &=\frac{-1\pm \sqrt{81}}{2}\\ \\ &=\frac{-1\pm 9}{2}\end{aligned}\] So \[x=-5\quad \text{or}\quad x=4\] 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 \[(-7)^2-20=29\] The value of the right-hand side is \[-1 \cdot -7=7\] So we have found for \(x<-5\) that \(x^2-20 > -x\).
Next we choose a value \(-5<x<4\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-20=-4\] The value of the right-handside is \[-1\cdot -4=4\] So we have found for \(-5<x<4\) that \(x^2-20 < -x\).
Finally we choose a value \(x>4\), say \(x=5\). The value of the left-hand side of the inequality is then \[(5)^2-20=5\] The value of the right-hand side is \[-1 \cdot 5=-5\] So we have found for \(x>4\) that \(x^2-20>-x\).

So we can conclude that \[x^2-20 > -x\] when \(x<-5\) or \(x>4\).