\(x<-1 \lor x > 2\)

We have the inequality \[x^2-2 > x\] but first we solve the following equation: \(x^2-2 = x \), that is \( x^2-x-2 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{1\pm \sqrt{(1)^2-4 \cdot 1 \cdot -2}}{2}\\ \\ &=\frac{1\pm \sqrt{9}}{2}\\ \\ &=\frac{1\pm 3}{2}\end{aligned}\] So \[x=-1\quad \text{or}\quad x=2\] Now we explore where the inequality is true.
First we take a value \(x<-1\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-2=7\] The value of the right-hand side is \[1 \cdot -3=-3\] So we have found for \(x<-1\) that \(x^2-2 > x\).
Next we choose a value \(-1<x<2\), say \(x=0\). The value of the left-hand side of the inequality is then \[(0)^2-2=-2\] The value of the right-handside is \[1\cdot 0=0\] So we have found for \(-1<x<2\) that \(x^2-2 < x\).
Finally we choose a value \(x>2\), say \(x=3\). The value of the left-hand side of the inequality is then \[(3)^2-2=7\] The value of the right-hand side is \[1 \cdot 3=3\] So we have found for \(x>2\) that \(x^2-2>x\).

So we can conclude that \[x^2-2 > x\] when \(x<-1\) or \(x>2\).