Let $z_1$ and $z_2$ be two roots of the equation $z^2+az+b=0,$ z being complex. Further, assumed that the origin $z_1$ and $z_2$ from an equilateral triangle. Then

 The correct option is $a^2=3b$  
We have $z_1+z_2=-a,z_1{z}_2=b$  
$z_2=z_1{e}^{i\pi /3}$ 
 $z_2=z_1(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})=z_1(\frac{1}{2}+\frac{i\sqrt{3}}{2})$
$2z_2-z_1=\sqrt{3}i{z}_1$
${(2z_2-z_1)}^2=-3z_1^2$ 
Hence, $z_1^2+z_2^2=z_1{z}_2$
So, $a^2-2b=b$ 
$a^2=3b$