How can I prove that all quadratic equations are not injective? I was trying to prove that any quadratic formula ($ax^2 + bx + c$) will not be injective, but I have a little problem.
I started by assuming $f(x) = f(y)$.
We can put x and y into the general form of quadratic function, and we get
$ax^2 + bx + c = ay^2 + by + c$
Subtract c from both sides and organise a little bit and you get
$a(x^2-y^2) + b(x-y)=0$
$a(x+y)(x-y) + b(x-y)=0$
Here we can assume that $x\ne y$, hence $x-y\ne 0$, so divide both sides by x-y, and we get
$a(x+y)+b=0$
However, from here I couldn't find any contradictions, which is a problem because there must be a contradiction as quadratic functions are not injective. Can anyone tell me what the contradiction is here?

This is good stuff so far. You now know that you get $x = y$ if you have $a(x + y) + b \neq 0$, in which case, there is no contradiction. So, let's focus on when $a(x + y) + b = 0$. Can you find two values of $x$ and $y$, that are not equal to each other, but for which $a(x + y) + b = 0$? Or, equivalently, $x + y = -\frac{b}{a}$?
There are many possible answers to this, but here's one: we can let $x = -\frac{b}{2a} + 1$ and $y = -\frac{b}{2a} - 1$. These values of $x$ and $y$ could well contradict the definition of injectivity. We have
\begin{align*}
&a\left(-\frac{b}{2a} + 1\right)^2 + b\left(-\frac{b}{2a} + 1\right) + c \\
= \, &a\left(\frac{b^2}{4a} - 2\frac{b}{2a} + 1\right) - \frac{b^2}{2a} + b + c \\
= \, &\frac{ab^2}{4a} - \frac{2ab}{2a} + a - \frac{b^2}{2a} + b + c \\
= \, &\frac{b^2}{4} + a - \frac{b^2}{2a} + c.
\end{align*}
Try substituting in $-\frac{b}{2a} - 1$ instead, and show that it comes to the same value.
