Been Given a solution that I don't quite understand would appreciate some extra explanation.

Suppose that $x\in Z $

Then
$$x^2-x = x^2 - 2 \* \frac 12 \* x + (\frac 12)^2 - (\frac 12)^2$$

$$ = (x-\frac 12)^2 - \frac 14$$
$$ \ge -\frac 14$$

Thus $x^2-x$ is an integer larger or equals $-\frac 14$ and so $x^2 -x \ge 0$