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

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 \[x=-4\quad \text{or}\quad x=3\] Now we explore where the inequality is true.
First we take a value \(x<-4\), say \(x=-6\). The value of the left-hand side of the inequality is then \[(-6)^2-12=24\] The value of the right-hand side is \[-1 \cdot -6=6\] So we have found for \(x<-4\) that \(x^2-12 > -x\).
Next we choose a value \(-4<x<3\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-12=-3\] The value of the right-handside is \[-1\cdot -3=3\] So we have found for \(-4<x<3\) that \(x^2-12 < -x\).
Finally we choose a value \(x>3\), say \(x=4\). The value of the left-hand side of the inequality is then \[(4)^2-12=4\] The value of the right-hand side is \[-1 \cdot 4=-4\] So we have found for \(x>3\) that \(x^2-12>-x\).

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