\(x<-1 \lor x > 4\)

We have the inequality \[x^2-4 > 3 x\] but first we solve the following equation: \(x^2-4 = 3 x \), that is \( x^2-3 x-4 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{3\pm \sqrt{(3)^2-4 \cdot 1 \cdot -4}}{2}\\ \\ &=\frac{3\pm \sqrt{25}}{2}\\ \\ &=\frac{3\pm 5}{2}\end{aligned}\] So \[x=-1\quad \text{or}\quad x=4\] Now we explore where the inequality is true.
First we take a value \(x<-1\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-4=5\] The value of the right-hand side is \[3 \cdot -3=-9\] So we have found for \(x<-1\) that \(x^2-4 > 3 x\).
Next we choose a value \(-1<x<4\), say \(x=0\). The value of the left-hand side of the inequality is then \[(0)^2-4=-4\] The value of the right-handside is \[3\cdot 0=0\] So we have found for \(-1<x<4\) that \(x^2-4 < 3 x\).
Finally we choose a value \(x>4\), say \(x=5\). The value of the left-hand side of the inequality is then \[(5)^2-4=21\] The value of the right-hand side is \[3 \cdot 5=15\] So we have found for \(x>4\) that \(x^2-4>3 x\).

So we can conclude that \[x^2-4 > 3 x\] when \(x<-1\) or \(x>4\).