Can you find the maximum or minimum of an equation without calculus? Without using calculus is it possible to find provably and exactly the maximum value
or the minimum value of a quadratic equation 
$$ y:=ax^2+bx+c $$
(and also without completing the square)?
I'd love to know the answer.

This is  almost the same as completing the square but .. for giggles.
If $a = 0$ we know $y = xb + c$ will get "extreme" and "extreme" positive and negative values of $x$ so no max or minimum is possible.
So if $ax^2 + bx + c = a(x^2 + x b/a)+c := a(x^2 + b'x) + c$ So finding the max/min is simply a matter of finding the max/min of $x^2 + b'x$ and multiplying by $a$ and adding $c$.
In other words .... wolog $a = 1$ and $c = 0$.
Now we know $x^2 + bx$ has only a min as $x^2$ is positive and as $|x|$ increases the $x^2$ term "overpowers" the $bx$ term.  Or if $x > |b|/2$ then $(x+ h)^2 + b(x + h) = x^2 + bx +h(2x + b) + h^2 > 0$ so the expression has no max value. 
So we want to find the minimum of $x^ + b'x = x(x + b)$.  Let $y := x - b'/2$ then $x(x + b')=(y -b'/2)(y + b'/2)= y^2 - (b'^2/4)$.  As $y^2 \ge 0$ the min will occur when $y = 0$ or in other words, $x= b'/2 = b/2a$
So the max/min of $ax^2 + bx + c$ occurs at $x = b/2a$ and the max/min value is $b^2/4 + b^2/2a + c$
Okay, that really was the same thing as completing the square but it didn't feel like it so what the @@@@.
