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/
 A: Here is a proof that works for any ordered commutative ring:
Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring.  Assume that $(a+b+c)c<0$.  We show that $b^2-4ac>0$.
Solution:
First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $c\neq 0$.
Then form the matrix 
$$M = \left( \begin{array}{cc}
  2c & b \\
  b & 2a \\
 \end{array}\right)$$
whose determinant is precicely $4ac-b^2$, and the matrix 
    $$ S = \left( \begin{array}{cc}
  1 & 0 \\
  2 & 1 \\
 \end{array}\right)$$
whose determinant is $1$.  Now from the multiplicativity of the determinant $\det(SM) =\det(S)\det(M)=\det(M)$.  But 
    $$\det(SM) = \det
 \left( \begin{array}{cc}
  2c & b \\
  4c+b & 2(a+b) \\
 \end{array}\right) = 4c(a+b) - b^2-4bc.$$
But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
The last element is always non-positive and the inequality is strict, so $\det(M)=4ac-b^2<0$.
A: 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, $(a+b+c)c<0\,\Rightarrow\, b^2-4ac>0$.
A: Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c \lt 0 $ and $b^2-4ac \le 0$ are simultaneously true.
1) By assumption
$$ (a + b + c) c \lt 0 $$
2) Negation of goal: $ b^2 - 4ac \gt 0$
$$ b^2 - 4ac \le 0 $$
3) From 2 by addition
$$ b^2 \le 4ac $$
4) expand 1
$$ ac + bc + c^2 < 0 $$
5) From 4 by addition
$$ ac < -bc - c^2 $$
6) From 5 by multiplication
$$ 4ac < -4bc -4c^2 $$
7) transitivity $x \le y \lt z$ implies $x \lt z$
$$ b^2 \lt -4bc - 4c^2 $$
8) from 7 by addition.
$$ b^2 + 4bc + 4c^2 \lt 0 $$
9) (7) is a perfect square
$$ (b + 2c)^2 < 0 $$
10) By assumption, $a, b, c \in \mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.
$$ \bot $$
Therefore, $b^2 - 4ac \gt 0$ .
A: We have
$$ 
(b^2 - 4ac) + 4(a+b+c)c =  (b+2c)^2 \ge 0 \\
\implies b^2 - 4ac \ge - 4(a+b+c)c > 0
$$
To give proper credit: The above approach was found after reading
guest's answer
and is just a simplification of that solution.
