I'm trying to understand the general cubic formula so let's try applying it to an example:
Let's take:
$$(x-1)(x-2)(x-3) = 0$$ $$\implies x^3 - 6x^2 +11x -6 = 0$$
So the general cubic formula should admit the roots 1, 2 and 3.
So:
$$a = 1$$ $$b = -6$$ $$c = 11$$ $$d = -6$$
Therefore:
$$\Delta_0 = b^2 - 3ac = (-6)^2 - 3 \cdot 1 \cdot 11 = 36 - 33 = 3 $$ $$\Delta_1 = 2b^3 - 9abc + 27a^2d \\ = 2 (-6)^3 - 9 \cdot 1 \cdot (-6) \cdot 11 + 27 \cdot 1 \cdot (-6) \\ = -432 + 594 - 162 = 0$$
and so:
$$C = \sqrt[3]{{\Delta_1 \pm \sqrt{\Delta_1^2 - 4 \Delta_0^3}}\over{2}} = \sqrt[3]{0 \pm \sqrt{0 - 4 \cdot 27} \over 2} = \sqrt[3]{\pm \sqrt{-108} \over 2} \approx \sqrt[3]{\pm 5.19i}$$
Supposedly one of the three roots is:
$$x = - {{1} \over {3a}}(b + C + {\Delta_0 \over C}) = - {{1} \over {3}}(-6 + \sqrt[3]{5.19i} + {3 \over \sqrt[3]{5.19i}})$$
Is that equal to 1, 2 or 3 when simplified? How? Or have I made an error?
Note that the value $\sqrt[3]{5.19} \approx \sqrt 3$. In fact, if you didn't round it off, the original value in your working would equal exactly that.
Also note that since $(-i)^3 = i$, one of the cube roots of $i$ is $-i$.
So one possible value of your expression is $-\frac 13(-6 + \sqrt 3 (-i) + \frac{3}{\sqrt 3 (-i)}) = -\frac 13 (-6 -i\sqrt 3 + i\sqrt 3) = 2$, which is indeed one of the integer roots you expected.
I did not check your final expression, I am assuming it is correct. Nevertheless, I have shown that what you ended up with can be simplified to a possible value of $2$.