Since it's a cubic there must be at least one real zero. When we find it we can divide by the corresponding factor to give us a quadratic which can be solved by completing squares or using the formula.
First we look for rational zeroes, by taking the coefficients of the highest and lowest degree terms.
8 is the highest degree coefficient and 27 is the lowest.
8 has factors 1, 2, 4, 8 and 27 has 1, 3, 9, 27. This gives rational zeroes:
1, 3, 9, 27, 1/2, 3/2, 9/2, 27/2, 1/4, 3/4, 9/4, 27/4, 1/8, 3/8, 9/8, 27/8, in positive and negative form. That's 32 in all. We could work through all 32. We only need to find one. We can narrow the search by evaluating the polynomial so that if p(x) is the polynomial, we can find p(a) which has the opposite sign to p(b) by substituting x=a and x=b. When x=0, y=27 and for x>0 p(x)>0 because all the terms are positive.
Now let's turn to x<0 and try x=-1: -8+18-45+27=-8. So there's a change of sign between x=0 and x=-1. We can eliminate all but -½, -¼, -¾, -⅛, -⅜ because all other rational zeroes are less than -1.
Substituting each of these 5 possibilities:
p(-½)<0, p(-¼)>0, p(-¾)=0, so we have found the zero -¾. Next we use synthetic division to find the quadratic:
-¾ | 8 18 45 27
8 -6 -9 | -27
8 12 36 | 0 = 4(2x2+3x+9).
Note that the distributing 4 changes the factor x+¾ into 4x+3, so the cubic factorises:
(4x+3)(2x2+3x+9)
Solve 2x2+3x+9=0 by using the formula:
x=(-3±√9-72)/4=(-3±√-63)/4. No real solutions, but complex solutions are:
(-3±3i√7)/4
The three zeroes are -¾, -¾(1+i√7), -¾(1-i√7).