Through a simple mathematical substitution, I have stumbled upon an alternative formula for solving a quadratic equation:
$$x=\frac{2c}{-b \pm \sqrt{b^{2}-4ac}}$$
(Please refer to my formula derivation here; this is my first question, so cannot include images yet).
I am not aware if this has been mentioned elsewhere.
I was just curious for a formula that reduces to the linear solution ($x=-\frac{c}{b}$) for the standard quadratic $ax^2 + bx + c = 0$, if we allow the case $a=0$. The standard solution formula $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$crashes for $a=0$ because division by zero is not allowed.
This alternative formula $x=\frac{2c}{-b \pm \sqrt{b^{2}-4ac}}$covers the case $a=0$, wherein it reduces to the linear equation $(bx + c = 0)$ solution ($x=-\frac{c}{b}$) for the negative square root of the discriminant, but the positive square root of the discriminant needs to be ignored/is a problem in that case.
What are your thoughts? Can we arrive at a general formula for a quadratic equation which includes all the cases ($a=0$ and/or $b=0$ and/or $c=0$)?
Look at the homogeneous case where we are working with points $(x:y)$ on the complex projective line and consider the homogeneous equation $$ax^2+bxy+cy^2=0.$$ The solutions are $$(x:y)=(-b \pm\sqrt{b^2-4ac}:2a)$$ and this works as long as $a,b,c$ are not all $0.$