\(x<1 \lor x > 4\)

We have the inequality \[x^2+4 > 5 x\] but first we solve the following equation: \(x^2+4 = 5 x \), that is \( x^2-5 x+4 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{5\pm \sqrt{(5)^2-4 \cdot 1 \cdot 4}}{2}\\ \\ &=\frac{5\pm \sqrt{9}}{2}\\ \\ &=\frac{5\pm 3}{2}\end{aligned}\] So \[x=1\quad \text{or}\quad x=4\] Now we explore where the inequality is true.
First we take a value \(x<1\), say \(x=-1\). The value of the left-hand side of the inequality is then \[(-1)^2+4=5\] The value of the right-hand side is \[5 \cdot -1=-5\] So we have found for \(x<1\) that \(x^2+4 > 5 x\).
Next we choose a value \(1<x<4\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2+4=8\] The value of the right-handside is \[5\cdot 2=10\] So we have found for \(1<x<4\) that \(x^2+4 < 5 x\).
Finally we choose a value \(x>4\), say \(x=5\). The value of the left-hand side of the inequality is then \[(5)^2+4=29\] The value of the right-hand side is \[5 \cdot 5=25\] So we have found for \(x>4\) that \(x^2+4>5 x\).

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