Where does the Quadratic Formula come from?

The method of completing the square can be used to solve any quadratic equation, i.e. any equation that can be put in the form $ax^2+bx+c=0$ where $a\ne 0$ . This includes the very equation $ax^2+bx+c=0$ itself...

$$ax^2+bx+c = 0$$

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

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

$$a$$

$$x^2+\frac{b}{a}x+\frac{b^2}{4a^2} = \frac{b^2}{4a^2}-\frac{c}{a} x$$ and adding the result to both sides. Note that

$$\frac{1}{2}\cdot\frac{b}{a}=\frac{b}{2a} \left( \frac{b}{2a}\right)^2=\frac{b^2}{4a^2} .$$

$$\left(x+\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}-\frac{c}{a}$$ Factor the perfect square trinomial.

$$\left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}$$ Rewrite the right hand side as a single fraction.

$$x+\frac{b}{2a} = \pm\sqrt{\frac{b^2-4ac}{4a^2}}$$ Take the square root of both sides.

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

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

$$x = \frac{\pm\sqrt{b^2-4ac} \; - \; b}{2a}$$ Rewrite the right hand side as a single fraction.

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

The Quadratic Formula: The solutions of $ax^2+bx+c=0$ , where $a\ne 0$ , are:

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