x^3+2x^2-3x<=0 inequation

Given the inequality:
$$- 3 x + \left(x^{3} + 2 x^{2}\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- 3 x + \left(x^{3} + 2 x^{2}\right) = 0$$
Solve:
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = -3$$
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = -3$$
This roots
$$x_{3} = -3$$
$$x_{1} = 0$$
$$x_{2} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{3}$$
For example, let's take the point
$$x_{0} = x_{3} - \frac{1}{10}$$
=
$$-3 + - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$- 3 x + \left(x^{3} + 2 x^{2}\right) \leq 0$$
$$\left(\left(- \frac{31}{10}\right)^{3} + 2 \left(- \frac{31}{10}\right)^{2}\right) - \frac{\left(-31\right) 3}{10} \leq 0$$

-1271      
------ <= 0
 1000      

one of the solutions of our inequality is:
$$x \leq -3$$

 _____           _____          
      \         /     \    
-------•-------•-------•-------
       x3      x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -3$$
$$x \geq 0 \wedge x \leq 1$$