\(x<-5 \lor x > 6\)

We have the inequality \[x^2-30 > x\] but first we solve the following equation: \(x^2-30 = x \), that is \( x^2-x-30 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{1\pm \sqrt{(1)^2-4 \cdot 1 \cdot -30}}{2}\\ \\ &=\frac{1\pm \sqrt{121}}{2}\\ \\ &=\frac{1\pm 11}{2}\end{aligned}\] So \[x=-5\quad \text{or}\quad x=6\] Now we explore where the inequality is true.
First we take a value \(x<-5\), say \(x=-7\). The value of the left-hand side of the inequality is then \[(-7)^2-30=19\] The value of the right-hand side is \[1 \cdot -7=-7\] So we have found for \(x<-5\) that \(x^2-30 > x\).
Next we choose a value \(-5<x<6\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-30=-14\] The value of the right-handside is \[1\cdot -4=-4\] So we have found for \(-5<x<6\) that \(x^2-30 < x\).
Finally we choose a value \(x>6\), say \(x=7\). The value of the left-hand side of the inequality is then \[(7)^2-30=19\] The value of the right-hand side is \[1 \cdot 7=7\] So we have found for \(x>6\) that \(x^2-30>x\).

So we can conclude that \[x^2-30 > x\] when \(x<-5\) or \(x>6\).