2x^2-2x+1<=1 inequation

Given the inequality:
$$\left(2 x^{2} - 2 x\right) + 1 \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2 x^{2} - 2 x\right) + 1 = 1$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(2 x^{2} - 2 x\right) + 1 = 1$$
to
$$\left(\left(2 x^{2} - 2 x\right) + 1\right) - 1 = 0$$
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
$$a = 2$$
$$b = -2$$
$$c = 0$$
, then

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

(-2)^2 - 4 * (2) * (0) = 4

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

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

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

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

61     
-- <= 1
50     

but

61     
-- >= 1
50     

Then
$$x \leq 0$$
no execute
one of the solutions of our inequality is:
$$x \geq 0 \wedge x \leq 1$$

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