Alternative proof that if $a,b,c \in \mathbb{R}$ and $(a+b+c)c<0$ then $b^2-4ac>0$? A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$.  Show that $b^2-4ac>0$.  There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$.  Is there any proof that does not involve any calculus, including IVT?
The question emerged from a reddit post 
https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/

If all you know is completing the square:
Suppose $b^2-4ac\leq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.
By completing the square we can rewrite $ax^2+bx+c=a\left(\left(x+\frac{b}{2a}\right)^2+d^2\right)$ for some real $d$.
We have $c=\frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:
$$\begin{align}(a+b+c)c&=\left(a+b+\frac{b^2}{4a}+ad^2\right)\left(\frac{b^2}{4a}+ad^2\right)\\
&=\left(a+b+\frac{b^2}{4a}\right)\frac{b^2}{4a}+\left(ad^2\frac{b^2}{4a}+\left(a+b+\frac{b^2}{4a}\right)ad^2\right)+\left(ad^2\right)^2\\
&=\frac{b^2}{4}\left(1+\frac{b}{a}+\frac{b^2}{4a^2}\right)+d^2\left(a^2+ab+\frac{b^2}{2}\right)+\left(ad^2\right)^2\\
&=\frac{b^2}{4}\left(1+\frac{b}{2a}\right)^2+d^2\left(a+\frac{b}{2}\right)^2+\left(ad^2\right)^2\\
&\geq0\text{.}
\end{align}$$
Thus by contrapositive, $(a+b+c)c<0\,\Rightarrow\, b^2-4ac>0$.
