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

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

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