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

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 \[x=-2\quad \text{or}\quad x=3\] 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 \[(-4)^2-6=10\] The value of the right-hand side is \[1 \cdot -4=-4\] So we have found for \(x<-2\) that \(x^2-6 > x\).
Next we choose a value \(-2<x<3\), say \(x=-1\). The value of the left-hand side of the inequality is then \[(-1)^2-6=-5\] The value of the right-handside is \[1\cdot -1=-1\] So we have found for \(-2<x<3\) that \(x^2-6 < x\).
Finally we choose a value \(x>3\), say \(x=4\). The value of the left-hand side of the inequality is then \[(4)^2-6=10\] The value of the right-hand side is \[1 \cdot 4=4\] So we have found for \(x>3\) that \(x^2-6>x\).

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