$\forall x \in Z $ such that $ x^2 -x \ge 0$ 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 $x^2 -x \ge 0$
 A: The proof involves completing the square. Since the $\left(x-\displaystyle\frac{1}{2}\right)^2$ is always positive, the sum is bigger 
than $-\frac{1}{4}$. Since $x^2-x$ can only be an integer, the smallest the integer can be is 0.
A: If you want a more geometric proof, draw the graph of $y=x^2-x.$ It crosses the $x$-axis at $0$ and $1.$ It is below the axis for $0<x<1,$ and there are no integers in this range. So, $x^2-x \geq 0$ for all integers $x.$
A: There are more than a few ways to prove this, I'll warrant; completing the square of 
$x^2 - x$ is very clever and has already been well-explicated by our OP J2R5M3 and Chris Leary in his/her answer.  
I prefer a very straightforward approach based on elementary arithmetic:
Note that
$x^2 - x = x(x - 1), \; \forall x \in \Bbb Z; \tag 1$
we have
$x \le -1 \Longrightarrow x - 1 \le -1 \Longrightarrow x(x - 1) \ge 1; \tag 2$
$x = 0 \Longrightarrow x(x - 1) = 0; \tag 3$
$x \ge 1 \Longrightarrow x - 1 \ge 0 \Longrightarrow x(x - 1) \ge 0; \tag 4$
'twixt (2), (3) and (4) all cases are covered, and we see that
$\forall x \in \Bbb Z, \; x^2 - x = x(x - 1) \ge 0. \tag 5$
