I was going to go with the systems of equations approach, but got beaten to the chase. So here is a more "calculus" type answer.
Let the polynomial be $P(x)=ax^2+bx+c$. Then we can write the derivative as $P'(x)=2ax+b$.
Without loss of generality suppose that $a>0$. If it isn't then just multiply the polynomial by $-1$, which will not change its roots.
Note that $P'(x) > 0 $ when $x>-b/2a$ and $P'(x)<0$ when $x<-b/2a$. This means that $P$ is strictly increasing on $(-\infty,-b/2a)$ and then strictly decreasing on $(-b/2a,\infty)$. From this we should be able to conclude that it can only cross the $x-axis$ at most twice, but that is not yet a proof.
With the above in mind suppose that there are three roots $x_1,x_2$, and $x_3$. Then Rolle's theorem tells us that the derivative of $P$ is zero in the intervals $[x_1,x_2]$ and $[x_2,x_3]$. This is impossible because the derivative is linear and only has one root. Therefore there cannot be three roots.