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Suppose $A,B,C$ are real numbers, with $A\neq 0$ , and suppose $$ Ax^2+Bx+C=0 $$
Since $A$ is nonzero, we can divide by $A$ and obtain the equation $$ x^2 + bx+c = 0, $$ where $b=\frac{B}{A}$ and $c=\frac{C}{A}$ . This equation can be written as $$x^2 + bx + \frac{b^2}{4} -\frac{b^2}{4} +c =0,$$ so completing the square, i.e., applying the identity $(p+q)^2=p^2+2pq + q^2$ , yields $$\left(x+\frac{b}{2}\right)^2 = \frac{b^2}{4} - c.$$ Then, taking the
square root of both sides, and solving for $x$ , we obtain the solution formula \begin{eqnarray*} x &=& -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}-c}\nonumber\\ &=& \frac{B}{2A} \pm \sqrt{\frac{B^2}{4A^2}-\frac{C}{A}}\nonumber\\ &=& \frac{-B\pm\sqrt{B^2-4AC}}{2A}, \end{eqnarray*}and the derivation is completed.
A slightly less intuitive but more aesthetically pleasing approach to this derivation can be achieved by multiplying both sides of the equation
by $4a$ , resulting in the equation
in which the left-hand side can be expressed as $(2ax+b)^2$ . From here, the proof is identical.
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