Proof

Let $p(x)=ax^2 + bx +c$ be a polynomial over the field $F[X]$, i.e. where the coefficients $a,b,c$ are elements of some field $(F, + , \cdot).$
Let $a\neq 0$, i.e. $p$ be of degree $2$.

Since $a\neq 0$ and since $F$ is a field, there exists an $a^{-1}:=\frac 1a$ and we have $x^2+\frac bax +\frac ca = 0.$
Adding the term $\frac{b^2}{4a^2}$ on both sides of the equation gives us $x^2+\frac bax +\frac ca + \frac{b^2}{4a^2}= \frac{b^2}{4a^2}.$
The binomial theorem for $n=2$ shows that $\left(x+\frac{b}{2a}\right)^2+\frac ca=\frac{b^2}{4a^2}.$
Then $\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac ca=\frac{b^2-4ac}{4a^2}.$
This means that $x+\frac b{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac{\sqrt{b^2-4ac}}{2a}$.
Finally, this means that $x=\frac{-b \pm\sqrt{b^2-4ac}}{2a}.$

It follows that the formula $p(x)=0$ has exactly two roots given by $$x_1:=\frac{-b + \sqrt{b^2-4ac}}{2a},\quad x_2:=\frac{-b\; - \sqrt{b^2-4ac}}{2a}.$$
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