x^3-x-6=0 equation

Given the equation:
$$\left(x^{3} - x\right) - 6 = 0$$
transform
$$\left(- x + \left(x^{3} - 8\right)\right) + 2 = 0$$
or
$$\left(- x + \left(x^{3} - 2^{3}\right)\right) + 2 = 0$$
$$- (x - 2) + \left(x^{3} - 2^{3}\right) = 0$$
$$\left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right) - \left(x - 2\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(\left(x^{2} + 2 x\right) + 2^{2}\right) - 1\right) = 0$$
or
$$\left(x - 2\right) \left(x^{2} + 2 x + 3\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$x^{2} + 2 x + 3 = 0$$
This equation is of the form

a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 2$$
$$c = 3$$
, then

D = b^2 - 4 * a * c = 

(2)^2 - 4 * (1) * (3) = -8

Because D<0, then the equation
has no real roots,
but complex roots is exists.

x2 = (-b + sqrt(D)) / (2*a)

x3 = (-b - sqrt(D)) / (2*a)

or
$$x_{2} = -1 + \sqrt{2} i$$
$$x_{3} = -1 - \sqrt{2} i$$
The final answer for x^3 - x - 6 = 0:
$$x_{1} = 2$$
$$x_{2} = -1 + \sqrt{2} i$$
$$x_{3} = -1 - \sqrt{2} i$$