(x+11)*(2*x-5)/3<=0 inequation

Given the inequality:
$$\frac{\left(x + 11\right) \left(2 x - 5\right)}{3} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(x + 11\right) \left(2 x - 5\right)}{3} = 0$$
Solve:
Expand the expression in the equation
$$\frac{\left(x + 11\right) \left(2 x - 5\right)}{3} = 0$$
We get the quadratic equation
$$\frac{2 x^{2}}{3} + \frac{17 x}{3} - \frac{55}{3} = 0$$
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
$$a = \frac{2}{3}$$
$$b = \frac{17}{3}$$
$$c = - \frac{55}{3}$$
, then

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

(17/3)^2 - 4 * (2/3) * (-55/3) = 81

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

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

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

or
$$x_{1} = \frac{5}{2}$$
$$x_{2} = -11$$
$$x_{1} = \frac{5}{2}$$
$$x_{2} = -11$$
$$x_{1} = \frac{5}{2}$$
$$x_{2} = -11$$
This roots
$$x_{2} = -11$$
$$x_{1} = \frac{5}{2}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-11 + - \frac{1}{10}$$
=
$$- \frac{111}{10}$$
substitute to the expression
$$\frac{\left(x + 11\right) \left(2 x - 5\right)}{3} \leq 0$$
$$\frac{\left(- \frac{111}{10} + 11\right) \left(\frac{\left(-111\right) 2}{10} - 5\right)}{3} \leq 0$$

68     
-- <= 0
75     

but

68     
-- >= 0
75     

Then
$$x \leq -11$$
no execute
one of the solutions of our inequality is:
$$x \geq -11 \wedge x \leq \frac{5}{2}$$

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       x2      x1