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?

Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.
Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $\alpha$ and $\beta$. 
Now, $$~~~~~~~~~~~~~~~~~~~~~x-\alpha = x-\beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$ 
$$~~~~~~~~~~~~~~~~~~~~~k(x-\alpha)(x-\beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$
Equating equation 1 and 2 (k is a constant),
$$ax^{2}+b{x}+c=k(x-\alpha)(x-\beta)$$
$$ax^{2}+b{x}+c=k(x^{2}-\alpha x-\beta x+\alpha \beta)$$
$$ax^{2}+b{x}+c=kx^{2}-k(\alpha+\beta )x+k\alpha \beta)$$
Comparing both sides, we get $$a=k~;~b=-k(\alpha +\beta)~;~c=k\alpha \beta$$
From this, we get $$\alpha + \beta = \frac{-b}{a}~~;~~~\alpha \beta = \frac{c}{a}$$
Now, to get the value of $\alpha$, we follow the following procedure : 
First we take out the value of $\alpha - \beta$, so that we  can eliminate one term and find out the value of another.
$$(\alpha-\beta)^{2} = \alpha ^{2}+ \beta ^{2} - 2 \alpha \beta$$
Now we'll add $4 \alpha \beta $ on both the sides
$$(\alpha-\beta)^{2} +4 \alpha \beta  = \alpha ^{2}+ \beta ^{2} + 2 \alpha \beta$$
$$(\alpha-\beta)^{2} +4 \alpha \beta  = (\alpha + \beta )^{2} $$
$$(\alpha-\beta)^{2} = (\alpha + \beta )^{2} -4 \alpha \beta  $$
$$\alpha-\beta = \pm \sqrt{(\alpha + \beta )^{2} -4 \alpha \beta } $$
Substituting the values of $\alpha + \beta$ and $\alpha \beta$, we get,
$$\alpha-\beta = \pm \sqrt{(\frac{-b}{a} )^{2} -\frac{4c}{a} } $$
$$\alpha-\beta = \pm \sqrt{\frac{b^{2}-4ac}{a^{2}} } $$ or
$$~~~~~~~~~~~~~~~~~~~~~\alpha-\beta = \frac{\pm \sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$
Adding $Eq^{n} (2)~and~(3)$, we get,
$$2 \alpha = \frac{-b \pm \sqrt{b^{2}-4ac}}{a}$$
$$\alpha = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$
