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?

The proof of the quadratic formula takes advantage of completing the square. 
$$ax^2 + bx + c = 0, \ a \neq 0, \ a, \ b, \ c \ \in \mathbb R$$
$$ax^2 + bx = -c$$
$$x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$$
$$x^2 + \dfrac{b}{a}x + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$$
$$\left(x+\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \dfrac{b^2}{4a^2}$$
$$\left(x+\dfrac{b}{2a}\right)^2 = -\dfrac{4ac}{4a^2} + \dfrac{b^2}{4a^2}$$
$$\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{-4ac+b^2}{4a^2}$$
$$\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{b^2-4ac}{4a^2}$$
$$x+\dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2-4ac}{4a^2}}$$
$$x+\dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
$$x=-\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
$$\boxed{x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$$
This concludes the proof.
