The origin and the roots of the equation $x^2+ax+b=0$  form an equilateral triangle, if:

$x^2+ax+b=0$ $\Rightarrow \frac{-a\pm \sqrt{a^2-4b}}{2}$ 

 If $a^2-4b\ge 0$  then the three points, 0 and the two roots are real and so collinear. Therefore, considering $a^2-4b<0,$ we  have roots as

 $\frac{-a\pm i\sqrt{4b-a^2}}{2}$

Let $A=-\frac{a}{2}+i\frac{\sqrt{4b-a^2}}{2},B=-\frac{a}{2}-i\frac{\sqrt{4b-a^2}}{2}$

$OA=\sqrt{\frac{a^2}{4}+\frac{4b-a^2}{4}}=\sqrt{b}=OB$

$AB=\sqrt{\left(4b-a^2\right)}.$  Since $OA=OB=AB$

We get $4b-a^2=b$  i.e., $a^2=3b$