# quadratic equation in $\mathbb{C}$

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

 $\displaystyle ax^{2}\!+\!bx\!+\!c\;=\;0$ (1)

with real coefficients $a$, $b$, $c$ is valid as well for all complex values of these coefficients ($a\neq 0$), when the square root is determined as is presented in the parent entry (http://planetmath.org/TakingSquareRootAlgebraically).

Proof.  Multiplying (1) by $4a$ and adding $b^{2}$ to both sides gives an equivalent     (http://planetmath.org/Equivalent3) equation

 $4a^{2}x^{2}\!+\!4abx\!+\!4ac\!+\!b^{2}\;=\;b^{2}$

or

 $(2ax)^{2}\!+\!2\!\cdot\!2ax\!\cdot\!{b}\!+\!b^{2}\;=\;b^{2}\!-\!4ac$

or furthermore

 $(2ax\!+\!b)^{2}\;=\;b^{2}\!-\!4ac.$

Taking square root algebraically yields

 $2ax\!+\!b\;=\;\pm\!\sqrt{b^{2}\!-\!4ac},$

Note.  A quadratic formula is meaningful besides $\mathbb{C}$ also in other fields with characteristic   $\neq 2$  if one can find the needed “square root” (this may require a field extension).
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