The Galois group of \$ x ^ 4 + 3x ^ 3-3x-2 \$ on \$ mathbb {Q} \$.

I'm trying to find the Galois group $$x ^ 4 + 3x ^ 3-3x-2$$ more than $$mathbb {Q}$$ and it's stuck.

What I have tried is as follows.

Let $$p (x) = x ^ 4 + 3x ^ 3-3x-2$$. Note that the discriminant
begin {align *} D = -4 (3) 3 3 (-3) 3 3-27 (3) 4 4 (-2) 2 2-192 (3) (-3) (- 2) 2- 2-6 (3) 2 (-3) 22 (-2) +256 (-2) 33-27 (-3) 44 = -2183. end {align *}

Let $$p (x) = x ^ 4 + 3x ^ 3-3x-2$$. Note that, using variable $$x = y- frac {3} {4}$$, we have
begin {align *} q (y) = = y ^ 4 + frac {1} {8} (- 3 (3) ^ 2 + 8 (0)) y ^ 2 + frac {1} {8} (3 ^ 3-4 (3) (0) +8 (-3) y + frac {1} {256} (- 3 (3) ^ 4 + 16 (3) ^ 2 (0) -64 (3) (- 3) + 256 (-2)) \ & = y ^ 4- frac {27} {8} y ^ 2 + frac {3} {8} y- frac {179} {256}. end {align *}

And his solving cubic is
begin {align *} h (x) = x ^ 3 + frac {27} {4} x ^ 2 + frac {908} {64} x + frac {9} {64} end {align *}

The cubic resolution is complicated, so trying to check if the cubic resolution is reducible does not seem to be the right solution. So, it seems to me that I have only one index on the discriminant. Could any body help me?