(4x-2x^2-4)/(x^2+2x+1)<=0 inequation

Given the inequality:
$$\frac{\left(- 2 x^{2} + 4 x\right) - 4}{\left(x^{2} + 2 x\right) + 1} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(- 2 x^{2} + 4 x\right) - 4}{\left(x^{2} + 2 x\right) + 1} = 0$$
Solve:
Given the equation:
$$\frac{\left(- 2 x^{2} + 4 x\right) - 4}{\left(x^{2} + 2 x\right) + 1} = 0$$
Multiply the equation sides by the denominators:
1 + x^2 + 2*x
we get:
$$\frac{\left(\left(- 2 x^{2} + 4 x\right) - 4\right) \left(x^{2} + 2 x + 1\right)}{\left(x^{2} + 2 x\right) + 1} = 0$$
$$- 2 x^{2} + 4 x - 4 = 0$$
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:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -2$$
$$b = 4$$
$$c = -4$$
, then

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

(4)^2 - 4 * (-2) * (-4) = -16

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

x1 = (-b + sqrt(D)) / (2*a)

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

or
$$x_{1} = 1 - i$$
$$x_{2} = 1 + i$$
$$x_{1} = 1 - i$$
$$x_{2} = 1 + i$$
Exclude the complex solutions:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example

x0 = 0

$$\frac{-4 + \left(0 \cdot 4 - 2 \cdot 0^{2}\right)}{\left(0^{2} + 0 \cdot 2\right) + 1} \leq 0$$

-4 <= 0

so the inequality is always executed