\(x<-3 \lor x > 2\)

We have the inequality \[x^2-6 > -x\] but first we solve the following equation: \(x^2-6 = -x \), that is \( x^2+x-6 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{-1\pm \sqrt{(-1)^2-4 \cdot 1 \cdot -6}}{2}\\ \\ &=\frac{-1\pm \sqrt{25}}{2}\\ \\ &=\frac{-1\pm 5}{2}\end{aligned}\] So \[x=-3\quad \text{or}\quad x=2\] Now we explore where the inequality is true.
First we take a value \(x<-3\), say \(x=-5\). The value of the left-hand side of the inequality is then \[(-5)^2-6=19\] The value of the right-hand side is \[-1 \cdot -5=5\] So we have found for \(x<-3\) that \(x^2-6 > -x\).
Next we choose a value \(-3<x<2\), say \(x=-2\). The value of the left-hand side of the inequality is then \[(-2)^2-6=-2\] The value of the right-handside is \[-1\cdot -2=2\] So we have found for \(-3<x<2\) that \(x^2-6 < -x\).
Finally we choose a value \(x>2\), say \(x=3\). The value of the left-hand side of the inequality is then \[(3)^2-6=3\] The value of the right-hand side is \[-1 \cdot 3=-3\] So we have found for \(x>2\) that \(x^2-6>-x\).

So we can conclude that \[x^2-6 > -x\] when \(x<-3\) or \(x>2\).