Is this "alternate" Quadratic formula correct?

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Quadratic Function.

$$ax^2 + bx + c = y $$

Deriving formula for x:

$ax^2 + bx + c = y$

$x^2 + \frac{bx}{a} + \frac{c}{a} = \frac{y}{a}$

$x^2 + \frac{bx}{a} = \frac{y}{a} - \frac{c}{a}$

$x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{y}{a} - \frac{c}{a} + \frac{b^2}{4a^2}$

$(x + \frac{b}{2a})^2 = \frac{y}{a} - \frac{c}{a} + \frac{b^2}{4a^2} $

$(x + \frac{b}{2a})^2 = \frac{y}{a} + \frac{b^2-4ac}{4a^2} $

$(x + \frac{b}{2a})^2 = \frac{b^2+4ay-4ac}{4a^2} $

$x + \frac{b}{2a} = \sqrt{\frac{b^2+4ay-4ac}{4a^2}} $

$ x = \sqrt{\frac{b^2+4ay-4ac}{4a^2}} - \frac{b}{2a} $

$ x = \frac{-b +\sqrt{b^2+4ay-4ac}}{2a} $

$ x = \frac{-b - \sqrt{b^2+4ay-4ac}}{2a} $

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You don't need all that.

Regarding $y$ as constant, and assuming $a \ne 0$, just use the ordinary quadratic formula to solve for $x$ in the equation $$ax^2 + bx + (c-y)=0$$ which gives $$x = \frac{-b \pm \sqrt{b^2 - 4a(c-y)}}{2a}$$