Quadratic Formula for complex variable with real coefficients I have been upto proving the following
$$(\forall x\in \mathbb C,  ax^2 + b x + c = 0) \land(a\neq0)\Leftrightarrow \left(x = {\frac {-b \pm \sqrt{b^{2}-4ac}} {2a}}\right)$$
Due to equality we need to prove bi implication,i.e. $a\Rightarrow b$ and $a\Leftarrow b$.It involves just simple arithmetic to go from roots to quadratic equation, but the other way is bit tricky. This is what I have done so far
$$(\forall x\in \mathbb C,  ax^2 + b  x + c = 0) \Rightarrow \left(x = {\frac {-b \pm \sqrt{b^{2}-4ac}} {2a}}\right)$$
$$(\forall x\in \mathbb C,   c = -ax^2 - b  x ) \Rightarrow (2ax = {-b \pm \sqrt{b^{2}-4ac}} )$$
$$(\forall x\in \mathbb C,   c = -ax^2 - b  x ) \Rightarrow (2ax + b = {\pm \sqrt{b^{2}-4ac}} )$$
Using value of $c$ we get complete square, equation looks like this
$$(\forall x\in \mathbb C,   c = -ax^2 - b  x ) \Rightarrow ((2ax + b) = {\pm \sqrt{(2ax+b)^2} })$$
On this forum I have found discussion on forms like
$$(x  = {\pm \sqrt{x^2} })$$
For complex numbers, I had an impression that it is not true. But it raises question on the goal I am trying to proof.So, if I summarize, I need comments regarding 
1-is goal legal?
2-If legal then how to deal with resulting form? 
3- If not (1), then looking for some alternative approach to handle it.
thanks in advance  
 A: The trinomial
$$ax^2+bx+c=0$$
May be solved when $a,b,c$ are complex numbers, and it works fine with real $a,b,c$, since $\Bbb R\subset\Bbb C$.
However, we'll start with the real case.

Real resolution
We start with the equation
$$ax^2+bx+c=0$$
Where $a\neq0$, and $a,b,c\in\Bbb C$.
Then, rewrite the equation
$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$
Then, according to the sign of $\Delta=b^2-4ac$, there are three possibilities:


*

*If $\Delta=0$, the equation amounts to


$$\left(x+\frac{b}{2a}\right)^2=0$$
It has thus one double root, $x=-\frac{b}{2a}$.
But since $\Delta=0$, we can also write
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$


*

*If $\Delta>0$, then $\frac{b^2-4ac}{4a^2}$ is positive, so it's the square of number $\frac{\sqrt{\Delta}}{2a}$, and we can write


$$\left(x+\frac{b}{2a}\right)^2-\left(\frac{\sqrt{\Delta}}{2a}\right)^2=0$$
$$\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)=0$$
Hence
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$


*

*If $\Delta<0$, then the equation has no real solution, since it would amount to


$$\left(x+\frac{b}{2a}\right)^2=\frac{\Delta}{4a^2}$$
Where the left-hand side is $\geq0$, whereas the right-hand side is $<0$.
However, we can still write that $\Delta$ is the square of complex number $i\sqrt{4ac-b^2}$, hence we can write
$$\left(x+\frac{b}{2a}\right)^2-\left(\frac{i\sqrt{-\Delta}}{2a}\right)^2=0$$
$$\left(x+\frac{b}{2a}+\frac{i\sqrt{-\Delta}}{2a}\right)\left(x+\frac{b}{2a}-\frac{i\sqrt{-\Delta}}{2a}\right)=0$$
$$x=\frac{-b\pm i\sqrt{-\Delta}}{2a}$$
But, with the convention that $\sqrt{t}=i\sqrt{-t}$ when $t<0$, we can again write
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
So, in all cases, the solutions are
$$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
Hence your equivalence holds.

Complex resolution
Again, we start with the equation
$$ax^2+bx+c=0$$
Where $a\neq0$, and $a,b,c\in\Bbb C$.
Then, rewrite the equation
$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$
Now, a complex number $z$ has always a square root (two distinct ones if $z\neq0$, see for example here). We go on with a square root $\alpha$ of $b^2-4ac$, that is, $\alpha^2=b^2-4ac$, so that the equation can be written
$$\left(x+\frac{b}{2a}\right)^2-\frac{\alpha^2}{4a^2}=0$$
It's a difference of squares, and we know that $A^2-B^2=(A+B)(A-B)$, so
$$\left(x+\frac{b}{2a}+\frac{\alpha}{2a}\right)\left(x+\frac{b}{2a}-\frac{\alpha}{2a}\right)=0$$
So the roots are
$$x=-\frac{b}{2a}\pm\frac{\alpha}{2a}$$
Or if you prefer
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Important note: there is no canonical way to write that $\sqrt{t}$ is a specific square root of $t\in\Bbb C$, hence this symbol is meaningless. However, since it appears with a $\pm$, we will always work with both roots, so using the symbol here is harmless. Nevertheless, it should be avoided.
A: The cleanest way to show that is probably to set
$$z_{\pm} := \frac{-b\pm (b^2-4ac)^{\frac12}}{2a}$$
Where $a,b,c\in\mathbb C$ and see that
$$(z-z_+)(z-z_-) = az^2 + bz + c$$
And finally use the FTA to exclude the existence of any further roots.
