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

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 \[x=-4\quad \text{or}\quad x=5\] 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 \[(-6)^2-20=16\] The value of the right-hand side is \[1 \cdot -6=-6\] So we have found for \(x<-4\) that \(x^2-20 > x\).
Next we choose a value \(-4<x<5\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-20=-11\] The value of the right-handside is \[1\cdot -3=-3\] So we have found for \(-4<x<5\) that \(x^2-20 < x\).
Finally we choose a value \(x>5\), say \(x=6\). The value of the left-hand side of the inequality is then \[(6)^2-20=16\] The value of the right-hand side is \[1 \cdot 6=6\] So we have found for \(x>5\) that \(x^2-20>x\).

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