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
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