(x+5)*(x+8)<=0 inequation

Given the inequality:
$$\left(x + 5\right) \left(x + 8\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x + 5\right) \left(x + 8\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x + 5\right) \left(x + 8\right) = 0$$
We get the quadratic equation
$$x^{2} + 13 x + 40 = 0$$
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
$$a = 1$$
$$b = 13$$
$$c = 40$$
, then

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

(13)^2 - 4 * (1) * (40) = 9

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

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

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

or
$$x_{1} = -5$$
$$x_{2} = -8$$
$$x_{1} = -5$$
$$x_{2} = -8$$
$$x_{1} = -5$$
$$x_{2} = -8$$
This roots
$$x_{2} = -8$$
$$x_{1} = -5$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-8 + - \frac{1}{10}$$
=
$$- \frac{81}{10}$$
substitute to the expression
$$\left(x + 5\right) \left(x + 8\right) \leq 0$$
$$\left(- \frac{81}{10} + 5\right) \left(- \frac{81}{10} + 8\right) \leq 0$$

 31     
--- <= 0
100     

but

 31     
--- >= 0
100     

Then
$$x \leq -8$$
no execute
one of the solutions of our inequality is:
$$x \geq -8 \wedge x \leq -5$$

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