(x-8)(x+5)<0 inequation

Given the inequality:
$$\left(x - 8\right) \left(x + 5\right) < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 8\right) \left(x + 5\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x - 8\right) \left(x + 5\right) = 0$$
We get the quadratic equation
$$x^{2} - 3 x - 40 = 0$$
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
$$a = 1$$
$$b = -3$$
$$c = -40$$
, then

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

(-3)^2 - 4 * (1) * (-40) = 169

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

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

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

or
$$x_{1} = 8$$
$$x_{2} = -5$$
$$x_{1} = 8$$
$$x_{2} = -5$$
$$x_{1} = 8$$
$$x_{2} = -5$$
This roots
$$x_{2} = -5$$
$$x_{1} = 8$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$\left(x - 8\right) \left(x + 5\right) < 0$$
$$\left(-8 + - \frac{51}{10}\right) \left(- \frac{51}{10} + 5\right) < 0$$

131    
--- < 0
100    

but

131    
--- > 0
100    

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

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        /     \  
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       x2      x1