x-2/(x-5)(x-1)≥0 inequation

Given the inequality:
$$x - \frac{2}{x - 5} \left(x - 1\right) \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$x - \frac{2}{x - 5} \left(x - 1\right) = 0$$
Solve:
Given the equation:
$$x - \frac{2}{x - 5} \left(x - 1\right) = 0$$
Multiply the equation sides by the denominators:
-5 + x
we get:
$$\left(x - 5\right) \left(x - \frac{2}{x - 5} \left(x - 1\right)\right) = 0$$
$$x^{2} - 7 x + 2 = 0$$
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
$$a = 1$$
$$b = -7$$
$$c = 2$$
, then

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

(-7)^2 - 4 * (1) * (2) = 41

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

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

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

or
$$x_{1} = \frac{\sqrt{41}}{2} + \frac{7}{2}$$
$$x_{2} = \frac{7}{2} - \frac{\sqrt{41}}{2}$$
$$x_{1} = \frac{\sqrt{41}}{2} + \frac{7}{2}$$
$$x_{2} = \frac{7}{2} - \frac{\sqrt{41}}{2}$$
$$x_{1} = \frac{\sqrt{41}}{2} + \frac{7}{2}$$
$$x_{2} = \frac{7}{2} - \frac{\sqrt{41}}{2}$$
This roots
$$x_{2} = \frac{7}{2} - \frac{\sqrt{41}}{2}$$
$$x_{1} = \frac{\sqrt{41}}{2} + \frac{7}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(\frac{7}{2} - \frac{\sqrt{41}}{2}\right)$$
=
$$\frac{17}{5} - \frac{\sqrt{41}}{2}$$
substitute to the expression
$$x - \frac{2}{x - 5} \left(x - 1\right) \geq 0$$
$$- \frac{2}{-5 + \left(\frac{17}{5} - \frac{\sqrt{41}}{2}\right)} \left(-1 + \left(\frac{17}{5} - \frac{\sqrt{41}}{2}\right)\right) + \left(\frac{17}{5} - \frac{\sqrt{41}}{2}\right) \geq 0$$

                /       ____\     
                |12   \/ 41 |     
       ____   2*|-- - ------|     
17   \/ 41      \5      2   /     
-- - ------ - --------------- >= 0
5      2                ____      
                  8   \/ 41       
                - - - ------      
                  5     2         

but

                /       ____\    
                |12   \/ 41 |    
       ____   2*|-- - ------|    
17   \/ 41      \5      2   /    
-- - ------ - --------------- < 0
5      2                ____     
                  8   \/ 41      
                - - - ------     
                  5     2        

Then
$$x \leq \frac{7}{2} - \frac{\sqrt{41}}{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{7}{2} - \frac{\sqrt{41}}{2} \wedge x \leq \frac{\sqrt{41}}{2} + \frac{7}{2}$$

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