Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:

$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one get this?
Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?

Please help, Thanks!

Consider the quadratic $ax^2 + bx + c$. If $a = 0$, then finding the root is easy and boring, so we only look at the case when $a \neq 0$.
One way to get at the original quadratic formula is to complete the square. That is, note that
$$\begin{align}
 ax^2 + bx + c &= a\left(x^2 + \frac{b}{a} x + \frac{c}{a}\right) \\
&= a \left(x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a} \right) \\
&= a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right).
\end{align}$$
So when looking for solutions to $ax^2 + bx + c = 0$, we are looking at solutions to
$$ 0 = a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right) = a\left( x + \frac{b}{2a}\right)^2 + a\frac{4ac - b^2}{4a^2},$$
or equivalently
$$ \left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}.$$
Taking square roots and solving for $x$ gives the original quadratic formula.
Now let us derive the alternate form. Notice that if $c = 0$ then $x = 0$ is a root and $x = -b/a$ is the other root. So let us now consider $c \neq 0$. We now know that $x = 0$ is not a root. This means that if $r$ is a root of $ax^2 + bx + c = 0$, then $\frac{1}{r}$ will be a root of $a \frac{1}{x^2} + b \frac{1}{x} + c = 0$. Multiplying through by $x^2$ (which makes sense because we are not interested in $x = 0$), we are interested in reciprocals of the roots of
$$ a + bx + cx^2 = 0.$$
The original quadratic formula applies, and we know the roots are
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}.$$
Since we are interested in the reciprocals, we find the roots to be
$$ x = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}.$$
This isn't necessarily the best derivation, but I like it. $\diamondsuit$
