-2x^2-8x+6>0 inequation

Given the inequality:
$$\left(- 2 x^{2} - 8 x\right) + 6 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 2 x^{2} - 8 x\right) + 6 = 0$$
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
$$a = -2$$
$$b = -8$$
$$c = 6$$
, then

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

(-8)^2 - 4 * (-2) * (6) = 112

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

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

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

or
$$x_{1} = - \sqrt{7} - 2$$
$$x_{2} = -2 + \sqrt{7}$$
$$x_{1} = - \sqrt{7} - 2$$
$$x_{2} = -2 + \sqrt{7}$$
$$x_{1} = - \sqrt{7} - 2$$
$$x_{2} = -2 + \sqrt{7}$$
This roots
$$x_{1} = - \sqrt{7} - 2$$
$$x_{2} = -2 + \sqrt{7}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(- \sqrt{7} - 2\right) + - \frac{1}{10}$$
=
$$- \sqrt{7} - \frac{21}{10}$$
substitute to the expression
$$\left(- 2 x^{2} - 8 x\right) + 6 > 0$$
$$\left(- 2 \left(- \sqrt{7} - \frac{21}{10}\right)^{2} - 8 \left(- \sqrt{7} - \frac{21}{10}\right)\right) + 6 > 0$$

                      2              
114     /  21     ___\        ___    
--- - 2*|- -- - \/ 7 |  + 8*\/ 7  > 0
 5      \  10        /               
    

Then
$$x < - \sqrt{7} - 2$$
no execute
one of the solutions of our inequality is:
$$x > - \sqrt{7} - 2 \wedge x < -2 + \sqrt{7}$$

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