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

We have the inequality \[x^2+5 > 6 x\] but first we solve the following equation: \(x^2+5 = 6 x \), that is \( x^2-6 x+5 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{6\pm \sqrt{(6)^2-4 \cdot 1 \cdot 5}}{2}\\ \\ &=\frac{6\pm \sqrt{16}}{2}\\ \\ &=\frac{6\pm 4}{2}\end{aligned}\] So \[x=1\quad \text{or}\quad x=5\] 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 \[(-1)^2+5=6\] The value of the right-hand side is \[6 \cdot -1=-6\] So we have found for \(x<1\) that \(x^2+5 > 6 x\).
Next we choose a value \(1<x<5\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2+5=9\] The value of the right-handside is \[6\cdot 2=12\] So we have found for \(1<x<5\) that \(x^2+5 < 6 x\).
Finally we choose a value \(x>5\), say \(x=6\). The value of the left-hand side of the inequality is then \[(6)^2+5=41\] The value of the right-hand side is \[6 \cdot 6=36\] So we have found for \(x>5\) that \(x^2+5>6 x\).

So we can conclude that \[x^2+5 > 6 x\] when \(x<1\) or \(x>5\).