conditions an $a,b,c,d$ so that $z^2+az+b=0,z^2+cz+d=0$ are collinear .

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Find necessary and sufficient conditions an $a,b,c,d$ so that the equations $$z^2+az+b=0, \qquad z^2+cz+d=0$$ are collinear in the plane.

If I have understood the problem correctly I think the roots and coefficients are supposed to be complex.

I did try assuming the roots as $a+\iota b$, $c+\iota d$, $e+\iota f$ and $g+\iota h$. As they are collinear they must satisfy

$$\frac{b-d}{a-c}=\frac{f-g}{e-h}$$

I need some help moving forward.

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They’re collinear iff $\frac{a^2-4b}{c^2-4d}$ and $\frac{a^2-4b}{(a-c)^2}$ are positive reals.

a-c is twice the vector between the midpoints of both pairs. The determinants are the difference between the roots squared. Any complex number in the direction of a line, squared will give the same angle.