Quadratic Formula

Theorem: Roots of a quadratic equation $ax^2+bx+c=0$ with $a \neq 0$ are given by $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Prerequisites:
$(a+b)^2=a^2+b^2+2ab$ (proof)

Proof:

Given equation is:

$\qquad \quad ax^2+bx+c=0$

Dividing the whole equation by a:

$\qquad \quad x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$

Adding and subtracting $\dfrac{b^2}{4a^2}$:

$\qquad \quad \left(x^2+\dfrac{b}{a}x+\frac{b^2}{4a^2}\right)-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0$

Using $(a+b)^2=a^2+b^2+2ab$, equation can be rewritten as:

$\qquad \quad \left(x + \dfrac{b}{2a}\right)^2 -\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0\\ \Rightarrow\quad\;\;\begin{align}\left(x + \dfrac{b}{2a}\right)^2 &= \dfrac{b^2}{4a^2}-\dfrac{c}{a}\\ &= \dfrac{b^2 - 4ac}{4a^2}\end{align}$

Taking Square root on both sides:

$\Rightarrow\quad \left(x + \dfrac{b}{2a}\right)= \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}\\ \Rightarrow\quad\;\; x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence the result

Recommended:
Factor representation of polynomial
Sum and product of roots
Conditions for roots of a quadratic equation