How to derive the equation for x in a quadratic equation? 
Possible Duplicate:
Why can ALL quadratic equations be solved by the quadratic formula? 

How to derive this:
$x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)

The usual approach is to complete the square:
$$\begin{align*}
ax^2+bx+c&=a\left(x^2+\frac{b}ax+\frac{c}a\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}a\right)\\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\;,
\end{align*}$$
which equals $0$ if and only if $$a\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a}-c=\frac{b^2-4ac}{4a}\;.$$ Now divide both sides by $a$ to get
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\;,$$ and take square roots:
$$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac1{2a}\sqrt{b^2-4ac}\;.$$ Hence
$$x=-\frac{b}{2a}\pm\frac1{2a}\sqrt{b^2-4ac}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;.$$
