The origin and the roots of the equation x2+ax+b=0  form an equilateral triangle, if:

x2+ax+b=0 ⇒−a±a2−4b2  If a2−4b≥0  then the three points, 0 and the two roots are real and so collinear. Therefore, considering a2−4b<0, we  have roots as −a±i4b−a22Let A=−a2+i4b−a22, B=−a2−i4b−a22OA=a24+4b−a24=b=OBAB=(4b−a2).  Since OA=OB=ABWe get 4b-a2=b  i.e., a2=3b