6/(xsqrt(3)-3)+(x*sqrt(3)-6)/(x*sqrt(3)-9)>=2 inequation

Given the inequality:
$$\frac{6}{\sqrt{3} x - 3} + \frac{\sqrt{3} x - 6}{\sqrt{3} x - 9} \geq 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{6}{\sqrt{3} x - 3} + \frac{\sqrt{3} x - 6}{\sqrt{3} x - 9} = 2$$
Solve:
Given the equation:
$$\frac{6}{\sqrt{3} x - 3} + \frac{\sqrt{3} x - 6}{\sqrt{3} x - 9} = 2$$
Multiply the equation sides by the denominators:
-3 + x*sqrt(3) and -9 + x*sqrt(3)
we get:
$$\left(\sqrt{3} x - 3\right) \left(\frac{6}{\sqrt{3} x - 3} + \frac{\sqrt{3} x - 6}{\sqrt{3} x - 9}\right) = 2 \sqrt{3} x - 6$$
$$\frac{3 \left(x^{2} - \sqrt{3} x - 12\right)}{\sqrt{3} x - 9} = 2 \sqrt{3} x - 6$$
$$\frac{3 \left(x^{2} - \sqrt{3} x - 12\right)}{\sqrt{3} x - 9} \left(\sqrt{3} x - 9\right) = \left(\sqrt{3} x - 9\right) \left(2 \sqrt{3} x - 6\right)$$
$$3 x^{2} - 3 \sqrt{3} x - 36 = 6 x^{2} - 24 \sqrt{3} x + 54$$
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$3 x^{2} - 3 \sqrt{3} x - 36 = 6 x^{2} - 24 \sqrt{3} x + 54$$
to
$$- 3 x^{2} + 21 \sqrt{3} x - 90 = 0$$
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
$$a = -3$$
$$b = 21 \sqrt{3}$$
$$c = -90$$
, then

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

(21*sqrt(3))^2 - 4 * (-3) * (-90) = 243

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

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

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

or
$$x_{1} = 2 \sqrt{3}$$
$$x_{2} = 5 \sqrt{3}$$
$$x_{1} = 2 \sqrt{3}$$
$$x_{2} = 5 \sqrt{3}$$
$$x_{1} = 2 \sqrt{3}$$
$$x_{2} = 5 \sqrt{3}$$
This roots
$$x_{1} = 2 \sqrt{3}$$
$$x_{2} = 5 \sqrt{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2 \sqrt{3}$$
=
$$- \frac{1}{10} + 2 \sqrt{3}$$
substitute to the expression
$$\frac{6}{\sqrt{3} x - 3} + \frac{\sqrt{3} x - 6}{\sqrt{3} x - 9} \geq 2$$
$$\frac{-6 + \sqrt{3} \left(- \frac{1}{10} + 2 \sqrt{3}\right)}{-9 + \sqrt{3} \left(- \frac{1}{10} + 2 \sqrt{3}\right)} + \frac{6}{-3 + \sqrt{3} \left(- \frac{1}{10} + 2 \sqrt{3}\right)} \geq 2$$

                                     ___ /  1        ___\     
                              -6 + \/ 3 *|- -- + 2*\/ 3 |     
             6                           \  10          /     
--------------------------- + --------------------------- >= 2
       ___ /  1        ___\          ___ /  1        ___\     
-3 + \/ 3 *|- -- + 2*\/ 3 |   -9 + \/ 3 *|- -- + 2*\/ 3 |     
           \  10          /              \  10          /     

one of the solutions of our inequality is:
$$x \leq 2 \sqrt{3}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 2 \sqrt{3}$$
$$x \geq 5 \sqrt{3}$$