$\arg(\frac{z_1}{z_2})$ of complex equation If $z_1,z_2$ are the roots of the equation $az^2 + bz + c = 0$, with $a, b, c > 0$; $2b^2 > 4ac > b^2$; $z_1\in$ third quadrant; $z_2 \in$ second quadrant in the argand's plane then, show that $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$

$z_2 = \overline{z_1}\,$ since the quadratic has real coefficients, so $\,\arg\left(\dfrac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)=2 \arg(z_1)\,$.
Since $\,a,b,c \gt 0\,$ the roots are in the left half-plane $\,\operatorname{Re}(z_1)=\operatorname{Re}(z_2) = -\,\dfrac{b}{2a} \lt 0\,$, and given the condition that "$z_1\in$ third quadrant" $\,\operatorname{Im}(z_1) \lt 0\,$, so $\,z_1\,$ is the root with negative imaginary part:
$$z_1 = \dfrac{-b - i \sqrt{4ac - b^2}}{2a} = \sqrt{\dfrac{c}{a}}\left(-\dfrac{b}{2\sqrt{ac}} - i \sqrt{1 - \dfrac{b^2}{4ac}}\right)$$
The latter can be written as $\,z_1=\sqrt{\dfrac{c}{a}}\big(\cos(\varphi)+ i \sin(\varphi)\big)\,$ where $\,\varphi=\arg(z_1)\,$ and:
$$
\begin{cases}
\begin{align}
\cos(\varphi) &= -\dfrac{b}{2\sqrt{ac}} \\[5px]
\sin(\varphi) &= -\sqrt{1 - \dfrac{b^2}{4ac}}
\end{align}
\end{cases}
$$
Therefore $\,\varphi=2k\pi \pm \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right)\,$, and the $\,\sin(\varphi) \lt 0\,$ condition trims the solution set down to $\,\varphi=2k\pi \color{red}{-} \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right) = (2k-1)\pi + \arccos\left(\dfrac{b}{2\sqrt{ac}}\right)\,$, so in the end:
$$\,\arg\left(\dfrac{z_1}{z_2}\right)= 2 \varphi \;\;\equiv\;\; 2 \arccos\left(\dfrac{b}{2\sqrt{ac}}\right) = 2 \arccos\left(\sqrt{\dfrac{b^2}{4ac}}\right) \pmod{2 \pi} \,$$
