Solve using: complete quadratic equation? $$x^2 - x - 20 > 0$$
$$x^2 - x > 20$$
Step 3: 
So to complete it, let's use: $1/2$ so it ends up with $x$ as the middle term.
$$x^2 - x + 1/2 > 20 + 1/2$$
Step 4:
But then: 
It would be transform to:
$$(x - 1/2)^2 > 20 + 1/2$$
Step 5:
But resolving it, the initial inequation of "Step3" is different:
$$x^2 - 2(x)(1/2) + (-1/2)^2 > 20 + 1/2$$
Giving us:
$$x^2 - x + 1/4 > 20 + 1/2$$
Which is different than:
$$x^2 - x + 1/2 > 20 + 1/2$$
Any hint?

Given $$\displaystyle x^2-x-20=0\Rightarrow \underbrace{x^2-x+\left(\frac{1}{2}\right)^2}-\underbrace{20-\left(\frac{1}{2}\right)^2} =0$$
So we get $$\displaystyle \left(x-\frac{1}{2}\right)^2-\left(\frac{9}{2}\right)^2>0$$
So we get $$\displaystyle \left[x-\frac{1}{2}-\frac{9}{2}\right]\cdot \left[x-\frac{1}{2}+\frac{9}{2}\right]>0$$
So we get $$(x-5)(x+4)>0\Rightarrow x<-4\cup x>5$$
So our Solution is $$x\in \left(-\infty,-4\right)\cup (5,\infty)$$
