\(x<-3 \lor x > 4\)

We have the inequality \[x^2-12 > x\] but first we solve the following equation: \(x^2-12 = x \), that is \( x^2-x-12 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{1\pm \sqrt{(1)^2-4 \cdot 1 \cdot -12}}{2}\\ \\ &=\frac{1\pm \sqrt{49}}{2}\\ \\ &=\frac{1\pm 7}{2}\end{aligned}\] So \[x=-3\quad \text{or}\quad x=4\] 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 \[(-5)^2-12=13\] The value of the right-hand side is \[1 \cdot -5=-5\] So we have found for \(x<-3\) that \(x^2-12 > x\).
Next we choose a value \(-3<x<4\), say \(x=-2\). The value of the left-hand side of the inequality is then \[(-2)^2-12=-8\] The value of the right-handside is \[1\cdot -2=-2\] So we have found for \(-3<x<4\) that \(x^2-12 < x\).
Finally we choose a value \(x>4\), say \(x=5\). The value of the left-hand side of the inequality is then \[(5)^2-12=13\] The value of the right-hand side is \[1 \cdot 5=5\] So we have found for \(x>4\) that \(x^2-12>x\).

So we can conclude that \[x^2-12 > x\] when \(x<-3\) or \(x>4\).