Why can a quadratic equation have only 2 roots? It is commonly known that the quadratic equation $ax^2+bx+c=0$ has two solutions given by: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ But how do I prove that another root couldn't exist?

I think derivation of quadratic formula is not enough....
Suppose there are three distinct roots $x,y,z$. One has
$$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\Rightarrow\begin{cases}a(x+y)+b=0\\a(x+z)+b=0\end{cases}$$ It follows $$a(z-y)=0\Rightarrow z=y$$ which is a contradiction.
If you wrote the theorem out, it would look like this

THEOREM: Let $a,b,$ and $c$ be real numbers with $a \ne 0$. Then $ax^2+bx+c=0$ if and only if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. 

That means

if $ax^2+bx+c=0$, then $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ 

and

if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, then $ax^2+bx+c=0$.