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!

I would like to add some point to answer posted user @mixedmath.
For equation
$$
ax^2 + bx + c = 0
$$
roots are
$$
x = \frac {-b\ \pm \ \sqrt {b^2 - 4ac} }{2a}
$$
Lets put 
$$
y = \frac {1}{x}
$$
Then we get converted equation as,
$$
cy^2 + by + a = 0
$$
Equivalent roots are,
$$
y = \frac {-b\ \mp \ \sqrt {b^2 - 4ac} }{2c}
$$
And thus if you invert to get values of $x$ then those are,
$$
x = \frac{1}{y} = \frac{2c}{-b \mp \sqrt{b^2 -4ac}}
$$
Do note the difference between the signs $\pm$ and $\mp$.
You can verify that by trying to equate both sides as follows:
$$
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2c}{-b \mp \sqrt{b^2 - 4ac}}
$$
Lets take one sign at a time. Lets choose $+$ from left side, this $-$ from right. It becomes,
$$
\frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{2c}{-b - \sqrt{b^2 - 4ac}}
$$
$$
(-b)^2 - (\sqrt{b^2 -4ac})^2 = 4ac
$$
$$
b^2 - b^2 + 4ac = 4ac
$$
$$
4ac = 4ac
$$
Thus things are in correct place.
Lets verify with one quick example.
Consider a quadratic equation as follows,
$$
f(x) = (x-3)(2x-5) = 2x^2 - 11 x + 15 = 0
$$
According to first formula we get values as follows,
$$
x = \frac{11 \pm 1}{4}
$$
$$
\{x_+, x_-\} = \{3, \frac{5}{2}\}
$$
And with second formula we get,
$$
x = \frac{30}{11 \mp 1}
$$
$$
\{x_+, x_-\} = \{\frac{5}{2}, 3\}
$$
