3x-2(3+x)>-2 inequation

Given the inequality:
$$3 x - 2 \left(x + 3\right) > -2$$
To solve this inequality, we must first solve the corresponding equation:
$$3 x - 2 \left(x + 3\right) = -2$$
Solve:
Given the linear equation:

3*x-2*(3+x) = -2

Expand brackets in the left part

3*x-2*3-2*x = -2

Looking for similar summands in the left part:

-6 + x = -2

Move free summands (without x)
from left part to right part, we given:
$$x = 4$$
$$x_{1} = 4$$
$$x_{1} = 4$$
This roots
$$x_{1} = 4$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 4$$
=
$$\frac{39}{10}$$
substitute to the expression
$$3 x - 2 \left(x + 3\right) > -2$$
$$- 2 \left(3 + \frac{39}{10}\right) + \frac{3 \cdot 39}{10} > -2$$

-21      
---- > -2
 10      

Then
$$x < 4$$
no execute
the solution of our inequality is:
$$x > 4$$

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