x^2+2x-5<0 inequation

Given the inequality:
$$x^{2} + 2 x - 5 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$x^{2} + 2 x - 5 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = 2$$
$$c = -5$$
, then
$$D = b^2 - 4\ a\ c = $$
$$2^{2} - 1 \cdot 4 \left(-5\right) = 24$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = -1 + \sqrt{6}$$
Simplify
$$x_{2} = - \sqrt{6} - 1$$
Simplify
$$x_{1} = -1 + \sqrt{6}$$
$$x_{2} = - \sqrt{6} - 1$$
$$x_{1} = -1 + \sqrt{6}$$
$$x_{2} = - \sqrt{6} - 1$$
This roots
$$x_{2} = - \sqrt{6} - 1$$
$$x_{1} = -1 + \sqrt{6}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(- \sqrt{6} - 1\right) - \frac{1}{10}$$
=
$$- \sqrt{6} - \frac{11}{10}$$
substitute to the expression
$$x^{2} + 2 x - 5 < 0$$
$$2 \left(- \sqrt{6} - \frac{11}{10}\right) - 5 + \left(- \sqrt{6} - \frac{11}{10}\right)^{2} < 0$$

                     2              
  36   /  11     ___\        ___    
- -- + |- -- - \/ 6 |  - 2*\/ 6  < 0
  5    \  10        /               
    

but

                     2              
  36   /  11     ___\        ___    
- -- + |- -- - \/ 6 |  - 2*\/ 6  > 0
  5    \  10        /               
    

Then
$$x < - \sqrt{6} - 1$$
no execute
one of the solutions of our inequality is:
$$x > - \sqrt{6} - 1 \wedge x < -1 + \sqrt{6}$$

         _____  
        /     \  
-------ο-------ο-------
       x_2      x_1