-2(x-1)*8(x-2)≥0 inequation

Given the inequality:
$$8 \left(- 2 \left(x - 1\right)\right) \left(x - 2\right) \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$8 \left(- 2 \left(x - 1\right)\right) \left(x - 2\right) = 0$$
Solve:
Expand the expression in the equation
$$8 \left(- 2 \left(x - 1\right)\right) \left(x - 2\right) = 0$$
We get the quadratic equation
$$- 16 x^{2} + 48 x - 32 = 0$$
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
$$a = -16$$
$$b = 48$$
$$c = -32$$
, then

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

(48)^2 - 4 * (-16) * (-32) = 256

Because D > 0, then the equation has two roots.

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

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

or
$$x_{1} = 1$$
$$x_{2} = 2$$
$$x_{1} = 1$$
$$x_{2} = 2$$
$$x_{1} = 1$$
$$x_{2} = 2$$
This roots
$$x_{1} = 1$$
$$x_{2} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$8 \left(- 2 \left(x - 1\right)\right) \left(x - 2\right) \geq 0$$
$$8 \left(- 2 \left(-1 + \frac{9}{10}\right)\right) \left(-2 + \frac{9}{10}\right) \geq 0$$

-44      
---- >= 0
 25      

but

-44     
---- < 0
 25     

Then
$$x \leq 1$$
no execute
one of the solutions of our inequality is:
$$x \geq 1 \wedge x \leq 2$$

         _____  
        /     \  
-------•-------•-------
       x1      x2