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? 

Note that $$x_1^2 - 6x_1 = x_2^2 - 6x_2 \iff (x_1-x_2)(x_1+x_2)=6(x_1-x_2).$$ Suppose $x_1\ne x_2.$ So we have
$$x_1^2 - 6x_1 = x_2^2 - 6x_2 \iff (x_1-x_2)(x_1+x_2)=6(x_1-x_2)\iff x_1+x_2=6.$$ Now, $$6=x_1+x_2\ge 3+3=6.$$ Thus, the only possibility is $x_1=x_2(=3)$ which gives the desired contradiction.
