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

We have the inequality \[x^2-3 > 2 x\] but first we solve the following equation: \(x^2-3 = 2 x \), that is \( x^2-2 x-3 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{2\pm \sqrt{(2)^2-4 \cdot 1 \cdot -3}}{2}\\ \\ &=\frac{2\pm \sqrt{16}}{2}\\ \\ &=\frac{2\pm 4}{2}\end{aligned}\] 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 \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-3=6\] The value of the right-hand side is \[2 \cdot -3=-6\] So we have found for \(x<-1\) that \(x^2-3 > 2 x\).
Next we choose a value \(-1<x<3\), say \(x=0\). The value of the left-hand side of the inequality is then \[(0)^2-3=-3\] The value of the right-handside is \[2\cdot 0=0\] So we have found for \(-1<x<3\) that \(x^2-3 < 2 x\).
Finally we choose a value \(x>3\), say \(x=4\). The value of the left-hand side of the inequality is then \[(4)^2-3=13\] The value of the right-hand side is \[2 \cdot 4=8\] So we have found for \(x>3\) that \(x^2-3>2 x\).

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