Universal quadratic formula? Is there any way to write the quadratic formula such that it works for $ac= 0$ without having to make it piecewise?
The traditional solution of $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$ breaks when $a = 0$, and the less-traditional solution of $x = 2c / (-b \pm \sqrt{b^2 - 4ac})$ breaks when $c = 0$... so I'm wondering if there is a formula that works for both cases.
My attempt was to make the formula "symmetric" with respect to $a$ and $c$ by substituting $$x = y \sqrt{c/a}$$ to get $$y^{+1} + y^{-1} = -b/\sqrt{ac} = 2 w$$
whose solution is
$$y = -w \pm \sqrt{w^2 - 4}$$
which is clearly symmetric with respect to $a$ and $c$, but which doesn't really seem to get me anywhere if $ac = 0$.
(If this is impossible, it'd be nice if I could get some kind of theoretical explanation for it instead of a plain "this is not possible".)

$x=(−b±\sqrt{b^2−4ac})/2a ...(1)$
Take limit $a \rightarrow 0,$ apply L'Hospital rule and you should get 
$x = -c/b$ (solution of $bx + c = 0$) and another solution of infinity (neglecting since not defined).
There should not be any problem in equation (1) for the case of $c = 0$.
To find a symmetric formula for $ac$, both $a$ and $c$ must have equal significance in the quadratic equation which is not the case, that's why it's not possible.
