(√7-3)x>16-6√7 inequation

Given the inequality:
$$x \left(-3 + \sqrt{7}\right) > 16 - 6 \sqrt{7}$$
To solve this inequality, we must first solve the corresponding equation:
$$x \left(-3 + \sqrt{7}\right) = 16 - 6 \sqrt{7}$$
Solve:
Given the linear equation:

(sqrt(7)-3)*x = 16-6*sqrt(7)

Expand brackets in the left part

sqrt+7-3)*x = 16-6*sqrt(7)

Expand brackets in the right part

sqrt+7-3)*x = 16-6*sqrt7

Move free summands (without x)
from left part to right part, we given:
$$x \left(-3 + \sqrt{7}\right) + 3 = 19 - 6 \sqrt{7}$$
Divide both parts of the equation by (3 + x*(-3 + sqrt(7)))/x

x = 19 - 6*sqrt(7) / ((3 + x*(-3 + sqrt(7)))/x)

$$x_{1} = -3 + \sqrt{7}$$
$$x_{1} = -3 + \sqrt{7}$$
This roots
$$x_{1} = -3 + \sqrt{7}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(-3 + \sqrt{7}\right) + - \frac{1}{10}$$
=
$$- \frac{31}{10} + \sqrt{7}$$
substitute to the expression
$$x \left(-3 + \sqrt{7}\right) > 16 - 6 \sqrt{7}$$
$$\left(- \frac{31}{10} + \sqrt{7}\right) \left(-3 + \sqrt{7}\right) > 16 - 6 \sqrt{7}$$

/       ___\ /  31     ___\            ___
\-3 + \/ 7 /*|- -- + \/ 7 | > 16 - 6*\/ 7 
             \  10        /   

the solution of our inequality is:
$$x < -3 + \sqrt{7}$$

 _____          
      \    
-------ο-------
       x1