I will assume that $a\ne0$. Then, if $\beta =\dfrac{\,\overline b\,}{\overline a}$ and $\gamma=\dfrac{\,\overline c\,}{\overline a}$, the roots of $az^2+bz+c$ are the roots of $z^2+\beta z+\gamma$. If these roots are $r,s\in\Bbb R$, then\begin{align}z^2+\beta z+\gamma&=(z-r)(z-s)\\&=z^2-(r+s)z+rs\end{align}and therefore $\beta=-(r+s)$ and $\gamma=rs$. In particular, $\beta,\gamma\in\Bbb R$.
Now, note that\begin{align}(r-s)^2&=\bigl(-(r+s)\bigr)^2-4rs\\&=\beta^2-4\gamma\end{align}and therefore we must have $\beta^2-4\gamma\geqslant0$. On the other hand, if indeed $\beta,\gamma\in\Bbb R$ and if $\beta^2-4\gamma\geqslant0$, it follows from the quadratic formula that the roots of $z^2+\beta z+\gamma$ are indeed real.
So, the roots of $az^2+bz+c$ are real if and only if$$\dfrac{\,\overline b\,}{\overline a},\dfrac{\,\overline c\,}{\overline a}\in\Bbb R\quad\text{and}\quad\left(\dfrac{\,\overline b\,}{\overline a}\right)^2-4\dfrac{\,\overline c\,}{\overline a}\geqslant0.$$
Best Answer
Let $r$ be the purely imaginary root, so that $\overline{r} = -r$. We have these conjugate equations: $$a r^2 + b r^1 + c = 0 \qquad \text{and} \qquad \overline{a} r^2 - \overline{b} r^1 + \overline{c} = 0$$ Here's a neat trick: Move the exponents on $r$ to subscripts ... $$a r_2 + b r_1 + c = 0 \qquad \text{and} \qquad \overline{a}r_2 - \overline{b} r_1 + \overline{c} = 0$$ ... and solve the linear system for them ... $$ r_1 = \frac{a\overline{c}-\overline{a}c}{a\overline{b}+\overline{a}b} \qquad\qquad r_2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b} $$ (Dealing with the case of $a\overline{b}+\overline{a}b=0$ is left as an exercise to the reader.) Now, return the subscripts to their positions as exponents ... $$ r^1 = \frac{a\overline{c}-\overline{a}c}{a\overline{b}+\overline{a}b} \qquad\qquad r^2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b} $$ ... and simply observe the deep algebraic truth, $r^2 = (r^1)^2$ ... $$\left(\;\frac{a\overline{c} -\overline{a}c}{a\overline{b}+\overline{a}b}\;\right)^2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}$$ Squaring and clearing fractions, we have $$(a\overline{c} -\overline{a}c)^2 = -(a\overline{b}+\overline{a}b)( b\overline{c}+\overline{b}c)$$