3x^2-13x+14<0 inequation

Given the inequality:
$$3 x^{2} - 13 x + 14 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$3 x^{2} - 13 x + 14 = 0$$
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
$$a = 3$$
$$b = -13$$
$$c = 14$$
, then

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

(-13)^2 - 4 * (3) * (14) = 1

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

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

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

or
$$x_{1} = \frac{7}{3}$$
Simplify
$$x_{2} = 2$$
Simplify
$$x_{1} = \frac{7}{3}$$
$$x_{2} = 2$$
$$x_{1} = \frac{7}{3}$$
$$x_{2} = 2$$
This roots
$$x_{2} = 2$$
$$x_{1} = \frac{7}{3}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$3 x^{2} - 13 x + 14 < 0$$
$$- \frac{13 \cdot 19}{10} + 3 \left(\frac{19}{10}\right)^{2} + 14 < 0$$

 13    
--- < 0
100    

but

 13    
--- > 0
100    

Then
$$x < 2$$
no execute
one of the solutions of our inequality is:
$$x > 2 \wedge x < \frac{7}{3}$$

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