x^2-6x+1<0 inequation

Given the inequality:
$$\left(x^{2} - 6 x\right) + 1 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x^{2} - 6 x\right) + 1 = 0$$
Solve:
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
$$a = 1$$
$$b = -6$$
$$c = 1$$
, then

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

(-6)^2 - 4 * (1) * (1) = 32

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

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

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

or
$$x_{1} = 2 \sqrt{2} + 3$$
$$x_{2} = 3 - 2 \sqrt{2}$$
$$x_{1} = 2 \sqrt{2} + 3$$
$$x_{2} = 3 - 2 \sqrt{2}$$
$$x_{1} = 2 \sqrt{2} + 3$$
$$x_{2} = 3 - 2 \sqrt{2}$$
This roots
$$x_{2} = 3 - 2 \sqrt{2}$$
$$x_{1} = 2 \sqrt{2} + 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(3 - 2 \sqrt{2}\right)$$
=
$$\frac{29}{10} - 2 \sqrt{2}$$
substitute to the expression
$$\left(x^{2} - 6 x\right) + 1 < 0$$
$$\left(- 6 \left(\frac{29}{10} - 2 \sqrt{2}\right) + \left(\frac{29}{10} - 2 \sqrt{2}\right)^{2}\right) + 1 < 0$$

                     2               
  82   /29       ___\         ___    
- -- + |-- - 2*\/ 2 |  + 12*\/ 2  < 0
  5    \10          /                
    

but

                     2               
  82   /29       ___\         ___    
- -- + |-- - 2*\/ 2 |  + 12*\/ 2  > 0
  5    \10          /                
    

Then
$$x < 3 - 2 \sqrt{2}$$
no execute
one of the solutions of our inequality is:
$$x > 3 - 2 \sqrt{2} \wedge x < 2 \sqrt{2} + 3$$

         _____  
        /     \  
-------ο-------ο-------
       x2      x1