Proving that $f: [3, +\infty) \to \mathbb{R},\quad x \mapsto x^2 - 6x + 5$ is injective using contrapositive 
Given $$f: [3, +\infty) \to \mathbb{R},\quad x \mapsto x^2 - 6x + 5$$ determine whether $f$ is injective.

I'm trying to do this using contrapositive, i.e. 
$$(\forall x_1, x_2 \in [3, +\infty))(f(x_1) = f(x_2) \implies x_1 = x_2)$$ 
So, this yields me
$$x_1^2 - 6x_1 + 5 = x_2^2 - 6x_2 + 5$$
$$x_1^2 - 6x_1 = x_2^2 - 6x_2$$
$$x_1 + x_2 = 6.$$
Now, I'm stuck at the last step, what shall I do next? 

You can also prove it directly:
For $x \in (3, \infty)$ you have $f'(x) = 2x-6 >0$
Therefore $f(x)$ is strictly increasing and so injective on that interval.
The result holds on $[3, \infty)$ (because for $x \in (3, \infty)$ we have $f(x)>f(3)$)
