y³-2y²+y-2=0 equation

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

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

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = 1$$
, then

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

(0)^2 - 4 * (1) * (1) = -4

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

y2 = (-b + sqrt(D)) / (2*a)

y3 = (-b - sqrt(D)) / (2*a)

or
$$y_{2} = i$$
$$y_{3} = - i$$
The final answer for y^3 - 2*y^2 + y - 2 = 0:
$$y_{1} = 2$$
$$y_{2} = i$$
$$y_{3} = - i$$