x^2-17<0 inequation

Given the inequality:
$$x^{2} - 17 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$x^{2} - 17 = 0$$
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
$$a = 1$$
$$b = 0$$
$$c = -17$$
, then

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

(0)^2 - 4 * (1) * (-17) = 68

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

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

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

or
$$x_{1} = \sqrt{17}$$
$$x_{2} = - \sqrt{17}$$
$$x_{1} = \sqrt{17}$$
$$x_{2} = - \sqrt{17}$$
$$x_{1} = \sqrt{17}$$
$$x_{2} = - \sqrt{17}$$
This roots
$$x_{2} = - \sqrt{17}$$
$$x_{1} = \sqrt{17}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \sqrt{17} - \frac{1}{10}$$
=
$$- \sqrt{17} - \frac{1}{10}$$
substitute to the expression
$$x^{2} - 17 < 0$$
$$-17 + \left(- \sqrt{17} - \frac{1}{10}\right)^{2} < 0$$

                     2    
      /  1      ____\     
-17 + |- -- - \/ 17 |  < 0
      \  10         /     
    

but

                     2    
      /  1      ____\     
-17 + |- -- - \/ 17 |  > 0
      \  10         /     
    

Then
$$x < - \sqrt{17}$$
no execute
one of the solutions of our inequality is:
$$x > - \sqrt{17} \wedge x < \sqrt{17}$$

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