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.

We assume (for the sake of discovery; for this purpose it is good enough
if this is just an inspired guess)
that the curve $y = ax^2 + bx + c$ is symmetric around a vertical axis.
It's obvious this is true when $b = 0$, and if we have plotted
$y = ax^2 + bx + c$ for various other values of $a$, $b$, and $c$,
we may observe enough appearance of symmetry to suppose that it might be true in general.
So it's reasonable to say: supposing it were true, what would that tell
us about the minimum/maximum value of the polynomial?
We find the points on this curve of the form $(x,c)$ as follows:
\begin{align}
y &= c. \\
c &= ax^2 + bx + c. \\
0 &= ax^2 + bx = (ax + b)x.
\end{align}
Hence if $(x,c)$ is on the curve, then either $ax + b = 0$ or $x = 0$.
Setting $x_1 = -\dfrac ba$ and $x_2 = 0$, we can plug in these two values
for $x$ and confirm that indeed the two points
$\left(-\frac ba, c\right)$ and $(0, c)$ are on the curve.
Using the assumption that the curve is symmetric around a vertical axis,
the vertical axis would have to be halfway between
$\left(-\frac ba, c\right)$ and $(0, c)$, that is, it is
the line $x = -\dfrac b{2a}$.
If there is a global maximum or minimum, it is a reasonable guess that
it would be on this line, so let's see what we have at
$x_0 = -\dfrac b{2a}$. Plugging this into the equation and doing the
algebra to find the point $(x_0, y_0)$ on the curve,
\begin{align}
y_0 &= a\left(-\frac b{2a}\right)^2 + b\left(-\frac b{2a}\right) + c \\
&= c - \frac{b^2}{4a}.
\end{align}
So that's our candidate for the maximum or minimum value.
To prove this is correct, consider any value of $x$ other than 
$-\dfrac b{2a}$. Any such value can be expressed by its difference
from $-\dfrac b{2a}$, that is, we let
$$ x = -\frac b{2a} + t$$
where $t \neq 0$. Now plug this value into the equation
and do the algebra:
\begin{align}
y &= a\left(-\frac b{2a} + t\right)^2 + b\left(-\frac b{2a} + t\right) + c 
\\[.5ex]
&= at^2 + c - \frac{b^2}{4a}. \tag 1
\end{align}
If $a$ is positive, $at^2$ is positive, hence $y > c - \dfrac{b^2}{4a} = y_0$
for every point $(x,y)$ on the curve such that $x \neq x_0$,
and therefore $y_0 = c - \dfrac{b^2}{4a}$ is a minimum.
But if $a$ is negative, $at^2$ is negative, and similar reasoning
says that $y_0 = c - \dfrac{b^2}{4a}$ is a maximum.

Note that the proof made no assumption about the symmetry of the curve.
On the contrary, the equation $y = at^2 + c - \dfrac{b^2}{4a}$
can be used to prove that the curve is symmetric.
If we take this a little further, we can even derive the standard
quadratic formula from it.  The roots of the equation
$y = ax^2 + bx + c$ are the values of $x$ such that $y = 0$.
But as we know from Equation $(1)$, above, 
if we make the substitution $x = -\dfrac b{2a} + t$, that means
\begin{align}
0 = y &= ax^2 + bx + c \\ &= at^2 + c - \frac{b^2}{4a}. 
\end{align}
A little algebra (isolate the $at^2$ term on one side and divide by $a$)
gives us
$$
t^2 = \frac{b^2}{4a^2} - \frac ca. \tag 2
$$
In general, if $p^2 = q$ then $p = \pm \sqrt q$, so Equation $(2)$
tells us that
\begin{align}
t &= \pm \sqrt{\frac{b^2}{4a^2} - \frac ca} \\
 &= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}\\
 &= \pm \frac{\sqrt{b^2 - 4ac}}{\lvert 2a \rvert}\\
 &= \pm \frac{\sqrt{b^2 - 4ac}}{2a},
\end{align}
and recalling that we set $x = -\dfrac b{2a} + t$,
\begin{align}
x &= -\frac b{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\
  &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a},
\end{align}
which is precisely the usual quadratic formula.

Is the reasoning above actually just an example of "completing the square,"
or is it sufficiently different from the usual method of "completing the square" that it can be considered a different method?
The equation $x = -\dfrac b{2a} + t$ is equivalent to
$t = x  + \dfrac b{2a}$; the method of completing the square involves
expanding $\left(x  + \dfrac b{2a}\right)^2$;
and in fact we do see $t^2$ figuring prominently in the equations above.
Certainly we could be inspired to try completing the square after
noticing how neatly the equation
$ax^2 + bx + c = at^2 + c - \dfrac{b^2}{4a}$
simplified the problem; but we never actually expanded the 
binomial $\left(x  + \dfrac b{2a}\right)^2$, and we never subtracted
the original polynomial from it to find the amount we needed to
"complete" the square.
Instead, the quantity $c - \dfrac{b^2}{4a}$ just "appeared" in the
original equation as the result of a direct substitution.
I think that may be about as different from "completing the square"
as a purely algebraic method can get.
