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 :)
 A: 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
$$x=-\frac{b}{2a}\pm\frac1{2a}\sqrt{b^2-4ac}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;.$$
A: Since $a\neq0$ we can divide the equation by $a$ and have it as
$x^{2}+px+q=0$ where $p=\frac{b}{a},q=\frac{c}{a}$.
Note $$x^{2}+px+q=(x+\frac{p}{2})^{2}+q-\frac{p^{2}}{4}$$ so we need
to solve $(x+\frac{p}{2})^{2}+q-\frac{p^{2}}{4}=0$ 
so we have it
that $(x+\frac{p}{2})^{2}=q-\frac{p^{2}}{4}$. now take the square
root, reduce $\frac{p}{2}$ from both sides and substitute $p,q$. 
A: from $ax^2 + bx + c = 0 \Leftrightarrow  4a^2x^2 + 4abx + b^2 = b^2 - 4ac$.
Hence $(2ax + b)^2 = b^2 - 4ac$ then $2ax + b = \pm \sqrt{b^2 - 4ac}$.
from this, we have $ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
A: First off, note that you can easily transform the equation to $x^2-2dx+e=0$ by simply dividing out $a$ and a factor $-2$, then we're looking to explain the solution $x=d\pm\sqrt{d^2-e}$. 
Since $x^2-2dx+e=(x^2-2dx+d^2)-(d^2-e)=(x-d)^2-(d^2-e)=0$, taking the square root yields $x-d=\pm\sqrt{d^2-e}$, which was requested.
