Roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle Given that roots of $3z^2+3z+b$  and $0$ lie on equilateral triangle, find condition on $b$.
If $a$ is a  root then $a(\frac{1}{2}+i\frac{\sqrt 3 }{2})$ i also a root. Sum of root is $-1 = a(\frac{3}{2}+i\frac{\sqrt3}{2})$ and product is $a^2(\frac{1}{2}+i\frac{\sqrt 3 }{2}) = \frac{b}{3}$
I am asking is this correct and also does veeta's formula work in complex numbers? What is your method and how to proceed using this, find $a$ and put in second equation to get $b$ is complicated.
Thanks  a lot!!

Obviously the roots shall be complex, so that $12b>9$.
$$
z_{1,2}=\frac{-3\pm i\sqrt{12b-9}}{6}\Rightarrow |z_1|=|z_2|=\sqrt{\frac{b}{3}},\quad |z_2-z_1|=\frac{\sqrt{12b-9}}{3}\\
\Rightarrow 3b=12b-9\Rightarrow b=1.
$$
