Think about it geometrically $-$ then compute.
Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ has its apex at $(0,\alpha)$. Looking at the graph as it moves up and down you immediately see how the number of zeros depends on $\alpha$:
- for $\alpha>0$ we have no zeros.
- for $\alpha=0$ we have a single zero.
- for $\alpha<0$ we have two zeros.
Now we can introduct a second paramter $\beta$ to move the parabola left and right: $(x-\beta)^2+\alpha$ has its apex at $(\beta,\alpha)$.
Note: we used the fact that given a function $f(x)$, the graph of the function $f(x-\beta)$ looks exactly like the one of $f$ but shifted to the right by an amount $\beta$.
But of course, shifting a function left and right does not alter the amount of zeros. So it still only depends on $\alpha$. We expand the term a bit:
$$(x-\beta)^2+\alpha=x^2-2\beta x+\beta^2+\alpha.$$
Would your quadratic equation be given in this form, you would immediately see the amount of zeros as describes above. Unfortunately it is mostly given as
$$ x^2+\color{red}px+\color{blue}q=0$$
So instead, you have to look at what parts of the $\alpha$-$\beta$-form above corresponds to these new parameters $p$ and $q$:
$$ x^2\color{red}{-2\beta} x+\color{blue}{\beta^2+\alpha} = 0.$$
So we have $p=-2\beta$ and $q=\beta^2+\alpha$. If we only could extract $\alpha$ from these new parameters, we would immediately see the amount of zeros. But wait! We can!
$$\alpha=q-\beta^2=q-\left(\frac p2\right)^2.$$
This is exactly what you know as (the negative of) the discriminant.
I used the form $x^2+px+q=0$ and you used $ax^2+bx+c=0$. I hope this is not confusing you. Just divide by $a$ (if $a$ is non-zero):
$$x^2+ \color{red}{\frac ba}x+\color{blue}{\frac ca}=0$$
If you set $p=b/a$ and $q=c/a$ and plug this into my discriminant from above you obtain the one you know:
$$\left( \frac {\color{red}p}2 \right )^2-\color{blue}q = \frac{(\color{red}{b/a})^2}4-\color{blue}{\frac ca}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}.$$
Because $4a^2$ is always positive it suffices to look at $b^2-4ac$ as you did in your question.
Best Answer
With $z=x+iy$,
$$x^2+y^2+(A+B)x+i(A-B)y-Ci=0.$$
From the imaginary part, $y$ is unique when $$A\ne B.$$
Then
$$x^2+(A+B)x+y^2=0$$ requires $(A+B)^2=4y^2$ or
$$A+B=\pm2y=\pm\frac{2C}{A-B}$$
In other words,
$$A\ne B\land A^2-B^2=\pm2C.$$