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

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 \[x=-5\quad \text{or}\quad x=2\] 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 \[(-7)^2-10=39\] The value of the right-hand side is \[-3 \cdot -7=21\] So we have found for \(x<-5\) that \(x^2-10 > -3 x\).
Next we choose a value \(-5<x<2\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-10=6\] The value of the right-handside is \[-3\cdot -4=12\] So we have found for \(-5<x<2\) that \(x^2-10 < -3 x\).
Finally we choose a value \(x>2\), say \(x=3\). The value of the left-hand side of the inequality is then \[(3)^2-10=-1\] The value of the right-hand side is \[-3 \cdot 3=-9\] So we have found for \(x>2\) that \(x^2-10>-3 x\).

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