\(x<-1 \lor x > 6\)

We have the inequality \[x^2-6 > 5 x\] but first we solve the following equation: \(x^2-6 = 5 x \), that is \( x^2-5 x-6 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{5\pm \sqrt{(5)^2-4 \cdot 1 \cdot -6}}{2}\\ \\ &=\frac{5\pm \sqrt{49}}{2}\\ \\ &=\frac{5\pm 7}{2}\end{aligned}\] So \[x=-1\quad \text{or}\quad x=6\] 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 \[(-3)^2-6=3\] The value of the right-hand side is \[5 \cdot -3=-15\] So we have found for \(x<-1\) that \(x^2-6 > 5 x\).
Next we choose a value \(-1<x<6\), say \(x=0\). The value of the left-hand side of the inequality is then \[(0)^2-6=-6\] The value of the right-handside is \[5\cdot 0=0\] So we have found for \(-1<x<6\) that \(x^2-6 < 5 x\).
Finally we choose a value \(x>6\), say \(x=7\). The value of the left-hand side of the inequality is then \[(7)^2-6=43\] The value of the right-hand side is \[5 \cdot 7=35\] So we have found for \(x>6\) that \(x^2-6>5 x\).

So we can conclude that \[x^2-6 > 5 x\] when \(x<-1\) or \(x>6\).