(2x^2-5x+2)<=0 inequation

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

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

(-5)^2 - 4 * (2) * (2) = 9

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

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

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

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

8/25 <= 0

but

8/25 >= 0

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

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        /     \  
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       x2      x1