(x–5,7)(x–7,2)>0 inequation

Given the inequality:
$$\left(x - \frac{36}{5}\right) \left(x - \frac{57}{10}\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - \frac{36}{5}\right) \left(x - \frac{57}{10}\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x - \frac{36}{5}\right) \left(x - \frac{57}{10}\right) = 0$$
We get the quadratic equation
$$x^{2} - \frac{129 x}{10} + \frac{1026}{25} = 0$$
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
$$a = 1$$
$$b = - \frac{129}{10}$$
$$c = \frac{1026}{25}$$
, then

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

(-129/10)^2 - 4 * (1) * (1026/25) = 9/4

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

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

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

or
$$x_{1} = \frac{36}{5}$$
$$x_{2} = \frac{57}{10}$$
$$x_{1} = \frac{36}{5}$$
$$x_{2} = \frac{57}{10}$$
$$x_{1} = \frac{36}{5}$$
$$x_{2} = \frac{57}{10}$$
This roots
$$x_{2} = \frac{57}{10}$$
$$x_{1} = \frac{36}{5}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{57}{10}$$
=
$$\frac{28}{5}$$
substitute to the expression
$$\left(x - \frac{36}{5}\right) \left(x - \frac{57}{10}\right) > 0$$
$$\left(- \frac{36}{5} + \frac{28}{5}\right) \left(- \frac{57}{10} + \frac{28}{5}\right) > 0$$

4/25 > 0

one of the solutions of our inequality is:
$$x < \frac{57}{10}$$

 _____           _____          
      \         /
-------ο-------ο-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{57}{10}$$
$$x > \frac{36}{5}$$