\(x<1 \lor x > 3\)

We have the inequality \[x^2+3 > 4 x\] but first we solve the following equation: \(x^2+3 = 4 x \), that is \( x^2-4 x+3 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{4\pm \sqrt{(4)^2-4 \cdot 1 \cdot 3}}{2}\\ \\ &=\frac{4\pm \sqrt{4}}{2}\\ \\ &=\frac{4\pm 2}{2}\end{aligned}\] So \[x=1\quad \text{or}\quad x=3\] 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 \[(-1)^2+3=4\] The value of the right-hand side is \[4 \cdot -1=-4\] So we have found for \(x<1\) that \(x^2+3 > 4 x\).
Next we choose a value \(1<x<3\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2+3=7\] The value of the right-handside is \[4\cdot 2=8\] So we have found for \(1<x<3\) that \(x^2+3 < 4 x\).
Finally we choose a value \(x>3\), say \(x=4\). The value of the left-hand side of the inequality is then \[(4)^2+3=19\] The value of the right-hand side is \[4 \cdot 4=16\] So we have found for \(x>3\) that \(x^2+3>4 x\).

So we can conclude that \[x^2+3 > 4 x\] when \(x<1\) or \(x>3\).