Deriving the quadratic formula without completing the square

Consider the general quadratic equation

$ax^2+bx + c =0, a \ne 0.\quad\quad\quad(*)$

Let

$y = x + \omega \implies x = y-\omega.\quad\quad\quad(1)$

Substituting (1) into (*), we have

$a(y-\omega)^2+b(y-\omega)+c = 0.$

This is a new quadratic equation in $y$ :

$ay^2+(b-2a\omega)\cdot y + a\omega^2-b\omega+c=0.\quad\quad\quad(2)$

Let

$\omega = \frac{b}{2a} \quad\quad\quad(3)$

so that (2) becomes

$ay^2 + 0\cdot y + a(\frac{b}{2a})^2 - b(\frac{b}{2a}) +c=0.$

That is,

$ay^2 + \frac{-b^2}{4a} + c =0.\quad\quad\quad(4)$

Solving (4) for $y^2$ ,

$y^2 = \frac{b^2-4ac}{4a^2} \implies y = \pm \frac{\sqrt{b^2-4ac}}{2a}\quad\quad\quad(5)$

Substituting (3) and (5) into (1) gives

$x = \pm \frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a},$

i.e.,

$x = \frac{-b \pm\sqrt{b^2-4ac}}{2a}.$

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