u^2−3u+2>=0 inequation

Given the inequality:
$$\left(u^{2} - 3 u\right) + 2 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(u^{2} - 3 u\right) + 2 = 0$$
Solve:
This equation is of the form

a*u^2 + b*u + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$u_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$u_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -3$$
$$c = 2$$
, then

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

(-3)^2 - 4 * (1) * (2) = 1

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

u1 = (-b + sqrt(D)) / (2*a)

u2 = (-b - sqrt(D)) / (2*a)

or
$$u_{1} = 2$$
$$u_{2} = 1$$
$$u_{1} = 2$$
$$u_{2} = 1$$
$$u_{1} = 2$$
$$u_{2} = 1$$
This roots
$$u_{2} = 1$$
$$u_{1} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$u_{0} \leq u_{2}$$
For example, let's take the point
$$u_{0} = u_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$\left(u^{2} - 3 u\right) + 2 \geq 0$$
$$\left(- \frac{3 \cdot 9}{10} + \left(\frac{9}{10}\right)^{2}\right) + 2 \geq 0$$

 11     
--- >= 0
100     

one of the solutions of our inequality is:
$$u \leq 1$$

 _____           _____          
      \         /
-------•-------•-------
       u2      u1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$u \leq 1$$
$$u \geq 2$$