(x+3)(x-8)<_0 inequation

Given the inequality:
$$\left(x - 8\right) \left(x + 3\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 8\right) \left(x + 3\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x - 8\right) \left(x + 3\right) = 0$$
We get the quadratic equation
$$x^{2} - 5 x - 24 = 0$$
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
$$a = 1$$
$$b = -5$$
$$c = -24$$
, then

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

(-5)^2 - 4 * (1) * (-24) = 121

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

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

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

or
$$x_{1} = 8$$
$$x_{2} = -3$$
$$x_{1} = 8$$
$$x_{2} = -3$$
$$x_{1} = 8$$
$$x_{2} = -3$$
This roots
$$x_{2} = -3$$
$$x_{1} = 8$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-3 + - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$\left(x - 8\right) \left(x + 3\right) \leq 0$$
$$\left(-8 + - \frac{31}{10}\right) \left(- \frac{31}{10} + 3\right) \leq 0$$

111     
--- <= 0
100     

but

111     
--- >= 0
100     

Then
$$x \leq -3$$
no execute
one of the solutions of our inequality is:
$$x \geq -3 \wedge x \leq 8$$

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