Geometry, finding the possible values of 'a' $P (a,4)$, $Q (2,3)$, $R (3,-1)$ and $S (-2,4)$ are four points.
If $|PQ| = |RS|$, find the possible values of a
I know this is a pretty basic problem but I'm having a lot of trouble with it, here is my answer:
I found $|RS|$ to be $\sqrt{50}$  so |PQ| = sqrt-2a^2 + 1 = sqrt50
and from there I got $\sqrt{8}$ which is incorrect because that is only one value. 
The true answer should be; $-5$ or $9$ (according to textbook)
What is the proper method that I should use to come to the correct solution?

You know that
$$ \sqrt{50}= \lvert RS \rvert = \lvert PQ\rvert = \sqrt{(a-2)^2+(4-3)^2} = \sqrt{a^2-4a+4+1} = \sqrt{a^2-4a+5},$$
so a good starting point would be to find the $a$ so that $50 = a^2-4a+5$.
