\(x<-5 \lor x > 1\)

We have the inequality \[x^2-5 > -4 x\] but first we solve the following equation: \(x^2-5 = -4 x \), that is \( x^2+4 x-5 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{-4\pm \sqrt{(-4)^2-4 \cdot 1 \cdot -5}}{2}\\ \\ &=\frac{-4\pm \sqrt{36}}{2}\\ \\ &=\frac{-4\pm 6}{2}\end{aligned}\] So \[x=-5\quad \text{or}\quad x=1\] 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 \[(-7)^2-5=44\] The value of the right-hand side is \[-4 \cdot -7=28\] So we have found for \(x<-5\) that \(x^2-5 > -4 x\).
Next we choose a value \(-5<x<1\), say \(x=-4\). The value of the left-hand side of the inequality is then \[(-4)^2-5=11\] The value of the right-handside is \[-4\cdot -4=16\] So we have found for \(-5<x<1\) that \(x^2-5 < -4 x\).
Finally we choose a value \(x>1\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2-5=-1\] The value of the right-hand side is \[-4 \cdot 2=-8\] So we have found for \(x>1\) that \(x^2-5>-4 x\).

So we can conclude that \[x^2-5 > -4 x\] when \(x<-5\) or \(x>1\).