\(x<-3 \lor x > 6\)

We have the inequality \[x^2-18 > 3 x\] but first we solve the following equation: \(x^2-18 = 3 x \), that is \( x^2-3 x-18 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{3\pm \sqrt{(3)^2-4 \cdot 1 \cdot -18}}{2}\\ \\ &=\frac{3\pm \sqrt{81}}{2}\\ \\ &=\frac{3\pm 9}{2}\end{aligned}\] So \[x=-3\quad \text{or}\quad x=6\] 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 \[(-5)^2-18=7\] The value of the right-hand side is \[3 \cdot -5=-15\] So we have found for \(x<-3\) that \(x^2-18 > 3 x\).
Next we choose a value \(-3<x<6\), say \(x=-2\). The value of the left-hand side of the inequality is then \[(-2)^2-18=-14\] The value of the right-handside is \[3\cdot -2=-6\] So we have found for \(-3<x<6\) that \(x^2-18 < 3 x\).
Finally we choose a value \(x>6\), say \(x=7\). The value of the left-hand side of the inequality is then \[(7)^2-18=31\] The value of the right-hand side is \[3 \cdot 7=21\] So we have found for \(x>6\) that \(x^2-18>3 x\).

So we can conclude that \[x^2-18 > 3 x\] when \(x<-3\) or \(x>6\).