x^2-15x+26≥0 inequation

Given the inequality:
$$x^{2} - 15 x + 26 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$x^{2} - 15 x + 26 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -15$$
$$c = 26$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 26 + \left(-15\right)^{2} = 121$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 13$$
Simplify
$$x_{2} = 2$$
Simplify
$$x_{1} = 13$$
$$x_{2} = 2$$
$$x_{1} = 13$$
$$x_{2} = 2$$
This roots
$$x_{2} = 2$$
$$x_{1} = 13$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$x^{2} - 15 x + 26 \geq 0$$
$$- \frac{15 \cdot 19}{10} + \left(\frac{19}{10}\right)^{2} + 26 \geq 0$$

111     
--- >= 0
100     

one of the solutions of our inequality is:
$$x \leq 2$$

 _____           _____          
      \         /
-------•-------•-------
       x_2      x_1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 2$$
$$x \geq 13$$