Hmmm. There is a trick to it. I say that because most of the time they are only going to ask you to do this with cubics that have simple roots.
There is a simple (sort-of) way and there is the correct way.

The correct way is very long winded but it will properly solve your roots for you. Its here:
http://www.1728.com/cubic2.htm There is a calculator there to solve ones for you - so you can check your working too.

The simple way: - based on a little guesswork then some working out.

1. We expect (for a cubic) that the end result will be in the form (x - p)(x - q)(x - r) = 0 (three of them because its a cubic)
This is three parts multiplied together - so if any of them are 0 the final result will be zero.
Therefore we are looking for three solutions where:
(x - p) = 0 and (x - q) = 0 and (x - r) = 0
we are trying to find p, q, and r

2. First we guess a root. Try x = 0 (why not) then try some other simple integers if that doesn't work
E.g. (using your example)
2(0) - 3(0) - 3(0) +2 = 0
I.e. 2 = 0 well thats no good. So 0 is not a root (not p, or q, or r)

3. so try 1
2 - 3 - 3 +2 = 0 wow it looks like it might work but the signs don't add up so we get -2 = 0
but it suggests maybe we should try x = -1
-2 -3 +3 +2 = 0 excellent 0 = 0 so x = -1 is a factor
we would say (x + 1) = 0 if we wrote it in the form above (x - p) = 0

4. so now we have found one factor. we could keep guessing or we could factor this root out and get a quadratic ax^2 + bx + c = 0 which we can solve using methods you already know
(that quadratic will give you two roots which are the remaining two factors)

5. finding the quadratic ax^2 + bx + c = 0
Take your function and divide it by your root (our root is (x + 1) = 0)
(2x^3 - 3x^2 - 3x + 2) / (x + 1) = ax^2 + bx + c
or refactoring it:
2x^3 - 3x^2 - 3x + 2 = (x + 1) (ax^2 + bx + c)
which becomes (right hand side changing)
2x^3 - 3x^2 - 3x + 2 = ax^3 + bx^2 +cx + ax^2 + bx + c
which simplifies to (only rhs changing)
2x^3 - 3x^2 - 3x + 2 = ax^3 + (a + b)x^2 +(b + c)x + c

Now we can see the similarity on both sides - there is one matching pair for each kind of x.
We are going to use this similarity to do our work for us.
so:
for x^3 implies 2 = a
for x^2 implies -3 = a+b
for x implies -3 = b+c
for const implies 2 = c

so substituting what we know into those equations (from the bottom up):
so c = 2 therefore b = -3 - c = -3 - 2 = -5 (from the x implies line)
and a = -3 - b = -3 + 5 = 2 (from the X^2 implies line - which we already know from the x^3 line but it checks our working in case we make a simple error)
We could have worked it out in any order. We knew a and c in this case but sometimes you might only know one of them. Any of the equations will work to find the answer for a, b, c
so a = 2, b = -5, c = 2

so therefore your quadratic is 2x^2 - 5x + 2 = 0

6. Solve the quadratic using other methods
if you solve that you get two roots (helper link below for quadratic solver)
x = 2 and x = 0.5

which gives you your answers in agreement with your stated answer above
first we found (x +1) = 0 or X = -1
then we found X = 2 and X = 1/2

In summary
state what you need to find.
guess a root until you find one.
simplify the cubic to a quadratic.
solve that
gather your roots and present the answer

Here are two good videos which i liked. Hope it helps.

All simple methods of solution failed, therefore, I used a much more advanced method, found here: http://www.1728.com/cubic2.htm

Also, I used quadratic equation: x=(-b(+-)sqrt(b^2-4ac))/2a

Substitute in and get the roots.