Multiply through by x:
x3-2x2-x=3,
x3-2x2-x-3=0, which has only one real root.
There are various ways of finding the root. First note that when x=2, x3-2x2-x-3=-5 and when x=3, it's 3, so the root is between 2 and 3, because the sign changes from negative to positive.
Newton's iterative method can be used to find x. First we need the derivative of the expression: 3x2-4x-1.
xn+1=xn-(xn3-2xn2-xn-3)/(3xn2-4xn-1) and we start with x0=2
x1=11/3, x2=3.0285..., x3=2.7920..., x4=2.75796..., x5=2.757279..., x6=2.757278921.
SOLUTION: x=2.757278921 approx.
There are two complex roots. To find them we need to divide by the discovered real root.
The result of the division is x2+(x6-2)x+(x62-2x6-1)=0, which can be solved using the quadratic formula:
x=(2-x6±√((x6-2)2-4(x62-2x6-1)))/2=(-0.7573±i√3.7786)/2=(-0.7573±1.9439i)/2.
So x=-0.3786±0.9719i are the other two roots. (Figures rounded to 4 decimal places.)