Pedro,

I don't blame you; while there is a "cubic formula" that will grind out the solutions, it's a bit of a bear and most people try to avoid using it.

I suggest the following.

1. Plot the polynomial for a few easy values, say
   
   -3, -2, …, 3
   
   (plus-or-minus 3 because 3 is about 17/6, beyond which the cubic term will probably dominate, as the linear & constant terms are not big.)
   
   
2. If you hit a root a, divide out x-a and use the quadratic formula on what if left. [If you have a remainder you made a mistake]
   
   
3. Otherwise if p(n) and p(n+1) are on opposite sides of zero try simple fraction values between n and n+1. Or obtain an accurate graph using a graphing calculator and guess the zeros.

(1) Plot the polynomial for a few easy values, say

Other observations (think about these carefully)

*There is a negative root, because
p(-N)<0 for big N, p(0) > 0, and p(N)>0. Thus the graph crosses the x axis between -N and 0. If there is only one real root this is it.

* The sum of the roots is 17/6 and the product is -12/6. [Why? Expand (x-a)(x-b)(x-c) and look at the coefficients!] Thus if there are three real roots two are positive. This might make positive values a good place to start looking.

Good Hunting!
RD