(x-5)*(x-12)>=0 inequation

Given the inequality:
$$\left(x - 12\right) \left(x - 5\right) \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 12\right) \left(x - 5\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x - 12\right) \left(x - 5\right) = 0$$
We get the quadratic equation
$$x^{2} - 17 x + 60 = 0$$
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
$$a = 1$$
$$b = -17$$
$$c = 60$$
, then

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

(-17)^2 - 4 * (1) * (60) = 49

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

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

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

or
$$x_{1} = 12$$
$$x_{2} = 5$$
$$x_{1} = 12$$
$$x_{2} = 5$$
$$x_{1} = 12$$
$$x_{2} = 5$$
This roots
$$x_{2} = 5$$
$$x_{1} = 12$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 5$$
=
$$\frac{49}{10}$$
substitute to the expression
$$\left(x - 12\right) \left(x - 5\right) \geq 0$$
$$\left(-12 + \frac{49}{10}\right) \left(-5 + \frac{49}{10}\right) \geq 0$$

 71     
--- >= 0
100     

one of the solutions of our inequality is:
$$x \leq 5$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 5$$
$$x \geq 12$$