2x^2-10x+8<0 inequation

Given the inequality:
$$2 x^{2} - 10 x + 8 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 x^{2} - 10 x + 8 = 0$$
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
$$a = 2$$
$$b = -10$$
$$c = 8$$
, then

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

(-10)^2 - 4 * (2) * (8) = 36

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

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

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

or
$$x_{1} = 4$$
Simplify
$$x_{2} = 1$$
Simplify
$$x_{1} = 4$$
$$x_{2} = 1$$
$$x_{1} = 4$$
$$x_{2} = 1$$
This roots
$$x_{2} = 1$$
$$x_{1} = 4$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$2 x^{2} - 10 x + 8 < 0$$
$$- \frac{9 \cdot 10}{10} + 2 \left(\frac{9}{10}\right)^{2} + 8 < 0$$

31    
-- < 0
50    

but

31    
-- > 0
50    

Then
$$x < 1$$
no execute
one of the solutions of our inequality is:
$$x > 1 \wedge x < 4$$

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