-3(2-x)+5(2x+2)-3x<0 inequation

Given the inequality:
$$- 3 x + \left(- 3 \left(2 - x\right) + 5 \left(2 x + 2\right)\right) < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- 3 x + \left(- 3 \left(2 - x\right) + 5 \left(2 x + 2\right)\right) = 0$$
Solve:
Given the linear equation:

-3*(2-x)+5*(2*x+2)-3*x = 0

Expand brackets in the left part

-3*2+3*x+5*2*x+5*2-3*x = 0

Looking for similar summands in the left part:

4 + 10*x = 0

Move free summands (without x)
from left part to right part, we given:
$$10 x = -4$$
Divide both parts of the equation by 10

x = -4 / (10)

$$x_{1} = - \frac{2}{5}$$
$$x_{1} = - \frac{2}{5}$$
This roots
$$x_{1} = - \frac{2}{5}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{2}{5} + - \frac{1}{10}$$
=
$$- \frac{1}{2}$$
substitute to the expression
$$- 3 x + \left(- 3 \left(2 - x\right) + 5 \left(2 x + 2\right)\right) < 0$$
$$\left(- 3 \left(2 - - \frac{1}{2}\right) + 5 \left(\frac{\left(-1\right) 2}{2} + 2\right)\right) - \frac{\left(-1\right) 3}{2} < 0$$

-1 < 0

the solution of our inequality is:
$$x < - \frac{2}{5}$$

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