\(x<-2 \lor x > 5\)

We have the inequality \[x^2-10 > 3 x\] but first we solve the following equation: \(x^2-10 = 3 x \), that is \( x^2-3 x-10 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{3\pm \sqrt{(3)^2-4 \cdot 1 \cdot -10}}{2}\\ \\ &=\frac{3\pm \sqrt{49}}{2}\\ \\ &=\frac{3\pm 7}{2}\end{aligned}\] So \[x=-2\quad \text{or}\quad x=5\] Now we explore where the inequality is true.
First we take a value \(x<-2\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-10=6\] The value of the right-hand side is \[3 \cdot -4=-12\] So we have found for \(x<-2\) that \(x^2-10 > 3 x\).
Next we choose a value \(-2<x<5\), say \(x=-1\). The value of the left-hand side of the inequality is then \[(-1)^2-10=-9\] The value of the right-handside is \[3\cdot -1=-3\] So we have found for \(-2<x<5\) that \(x^2-10 < 3 x\).
Finally we choose a value \(x>5\), say \(x=6\). The value of the left-hand side of the inequality is then \[(6)^2-10=26\] The value of the right-hand side is \[3 \cdot 6=18\] So we have found for \(x>5\) that \(x^2-10>3 x\).

So we can conclude that \[x^2-10 > 3 x\] when \(x<-2\) or \(x>5\).