Proof of the Quadratic formula

The Quadratic Formula used to solve equations of the form $$ax^2+bx+c=0 \,\!$$. It is:


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

Proof:

Dividing our quadratic equation by $$a\,\!$$ (which is allowed because $$a\,\!$$ is non-zero), we have


 * $$x^2 + \frac{b}{a} x + \frac{c}{a}=0$$

which is equivalent to


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

The equation is now in a form in which we can conveniently complete the square. To "complete the square" is to add a constant (i.e., in this case, a quantity that does not depend on $$x\,\!$$) to the expression to the left of "$$=\,\!$$", that will make it a perfect square trinomial of the form $$x^2+2xy+y^2\,\!$$. Since $$2xy\,\!$$ in this case is $$\frac{b}{a} x $$, we must have $$y = \frac{b}{2a}$$, so we add the square of $$\frac{b}{2a}$$ to both sides, getting


 * $$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}.$$

The left side is now a perfect square; it is the square of $$\left(x + \frac{b}{2a}\right)$$. The right side can be written as a single fraction; the common denominator is $$4a^2\,\!$$. We get


 * $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.$$

Taking square roots of both sides yields


 * $$\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac\ }}{|2a|}\Leftrightarrow$$$$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\  }}{2a}$$

Subtracting $$\frac{b}{2a}$$ from both sides, we get


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