Why can ALL quadratic equations be solved by the quadratic formula? In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.
Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?

I'll just let the algebra speak for itself.
For $ a \ \ne \ 0 $
If
$ax^2 = bx + c $
Then
$ x = \frac{b \ \pm \sqrt{b^2 \ + \ 4ac}}{2a} $
PROOF: 
$ 4aax^2 = 4abx + 4ac $
$(2ax)^2 + b^2 = 4abx + 4ac + b^2 $
$(2ax)^2 - 4abx + b^2 = b^2 + 4ac $
$ (2ax - b)^2 = b^2 + 4ac $
$ 2ax - b = \pm \sqrt{b^2 + 4ac} $
$2ax = b \pm \sqrt{b^2 + 4ac} $
$ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $
Q.E.D.
SHORTCUT
Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.
For $ a \ne 0 $
If
$ ax^2 + bx + c = 0 $
Then
$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Make the replacements in the 'if'
$ ax^2 + (-b)x + (-c) = 0 $
$ ax^2 = bx + c $
Make the replacements in the 'then'
$x = \frac{-(-b) \pm \sqrt{(-b)^2 - 4a(-c)}}{2a} $
$ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $
Q.E.D.
