<s>[INST] 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!
 [/INST] \begin{align}
   \dfrac{-b \pm \sqrt{b^2 -4ac} }{2a} 
      &= \dfrac{-b \pm \sqrt{b^2 -4ac} }{2a} 
         \cdot \frac{-b \mp \sqrt{b^2 -4ac}}{-b \mp \sqrt{b^2 -4ac}} \\
      &= \dfrac{b^2-(b^2-4ac)}{2a(-b \mp \sqrt{b^2 -4ac})} \\
      &= \dfrac{4ac}{2a(-b \mp \sqrt{b^2 -4ac})} \\
      &= \dfrac{2c}{-b \pm \sqrt{b^2 -4ac}} \\
\end{align}
</s>