Since none of ±3 and ±1 are roots, this equation cannot have rational roots.
Graphically there is a real root between -.5 and -.75 and interval halving will get a root at approximately a = -.674.
The complex roots can be approximated by dividing out the approximate real root and then using the quadratic formula.
Cardan's method for the general cubic reduces the general cubic to the form in which the current equation is given. Then further algebra produces expressions for the 3 roots, but they are quite complicated. I worked the Cardan formula through numerically and obtained a = -.673594.
