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

For a triangle formed by $z_1,z_2,z_3$ to be equilateral, the condition is:
$$z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$$
Here, one vertex is the origin ($z_3=0$). The condition reduces to:
$$z_1^2+z_2^2=z_1{z}_2$$
Adding $2z_1{z}_2$ to both sides:
$$z_1^2+z_2^2+2z_1{z}_2=3z_1{z}_2$$
$$(z_1+z_2{)}^2=3z_1{z}_2$$
From the quadratic equation $z^2+az+b=0$:Sum of roots $z_1+z_2=-a$ Product of roots $z_1{z}_2=b$ Substituting these:
$$(-a{)}^2=3b⟹a^2=3b$$