$x<1 \lor x > 5$

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 $x>5$.