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....

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 $ax^2+bx+c=0$.
