-6*x^2+13*x-5>=0 inequation

Given the inequality:
$$\left(- 6 x^{2} + 13 x\right) - 5 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 6 x^{2} + 13 x\right) - 5 = 0$$
Solve:
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
$$a = -6$$
$$b = 13$$
$$c = -5$$
, then

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

(13)^2 - 4 * (-6) * (-5) = 49

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

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

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

or
$$x_{1} = \frac{1}{2}$$
$$x_{2} = \frac{5}{3}$$
$$x_{1} = \frac{1}{2}$$
$$x_{2} = \frac{5}{3}$$
$$x_{1} = \frac{1}{2}$$
$$x_{2} = \frac{5}{3}$$
This roots
$$x_{1} = \frac{1}{2}$$
$$x_{2} = \frac{5}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{2}$$
=
$$\frac{2}{5}$$
substitute to the expression
$$\left(- 6 x^{2} + 13 x\right) - 5 \geq 0$$
$$-5 + \left(- 6 \left(\frac{2}{5}\right)^{2} + \frac{2 \cdot 13}{5}\right) \geq 0$$

-19      
---- >= 0
 25      

but

-19     
---- < 0
 25     

Then
$$x \leq \frac{1}{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{1}{2} \wedge x \leq \frac{5}{3}$$

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