\(x<1 \lor x > 6\)

We have the inequality \[x^2+6 > 7 x\] but first we solve the following equation: \(x^2+6 = 7 x \), that is \( x^2-7 x+6 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{7\pm \sqrt{(7)^2-4 \cdot 1 \cdot 6}}{2}\\ \\ &=\frac{7\pm \sqrt{25}}{2}\\ \\ &=\frac{7\pm 5}{2}\end{aligned}\] So \[x=1\quad \text{or}\quad x=6\] 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 \[(-1)^2+6=7\] The value of the right-hand side is \[7 \cdot -1=-7\] So we have found for \(x<1\) that \(x^2+6 > 7 x\).
Next we choose a value \(1<x<6\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2+6=10\] The value of the right-handside is \[7\cdot 2=14\] So we have found for \(1<x<6\) that \(x^2+6 < 7 x\).
Finally we choose a value \(x>6\), say \(x=7\). The value of the left-hand side of the inequality is then \[(7)^2+6=55\] The value of the right-hand side is \[7 \cdot 7=49\] So we have found for \(x>6\) that \(x^2+6>7 x\).

So we can conclude that \[x^2+6 > 7 x\] when \(x<1\) or \(x>6\).