x^2+6x/6-2x+3/2<=12 inequation

Given the inequality:
$$\left(- 2 x + \left(x^{2} + \frac{6 x}{6}\right)\right) + \frac{3}{2} \leq 12$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 2 x + \left(x^{2} + \frac{6 x}{6}\right)\right) + \frac{3}{2} = 12$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(- 2 x + \left(x^{2} + \frac{6 x}{6}\right)\right) + \frac{3}{2} = 12$$
to
$$\left(\left(- 2 x + \left(x^{2} + \frac{6 x}{6}\right)\right) + \frac{3}{2}\right) - 12 = 0$$
Expand the expression in the equation
$$\left(\left(- 2 x + \left(x^{2} + \frac{6 x}{6}\right)\right) + \frac{3}{2}\right) - 12 = 0$$
We get the quadratic equation
$$x^{2} - x - \frac{21}{2} = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = - \frac{21}{2}$$
, then

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

(-1)^2 - 4 * (1) * (-21/2) = 43

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

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

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

or
$$x_{1} = \frac{1}{2} + \frac{\sqrt{43}}{2}$$
$$x_{2} = \frac{1}{2} - \frac{\sqrt{43}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{43}}{2}$$
$$x_{2} = \frac{1}{2} - \frac{\sqrt{43}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{43}}{2}$$
$$x_{2} = \frac{1}{2} - \frac{\sqrt{43}}{2}$$
This roots
$$x_{2} = \frac{1}{2} - \frac{\sqrt{43}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{43}}{2}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(\frac{1}{2} - \frac{\sqrt{43}}{2}\right) + - \frac{1}{10}$$
=
$$\frac{2}{5} - \frac{\sqrt{43}}{2}$$
substitute to the expression
$$\left(- 2 x + \left(x^{2} + \frac{6 x}{6}\right)\right) + \frac{3}{2} \leq 12$$
$$\frac{3}{2} + \left(\left(\frac{6 \left(\frac{2}{5} - \frac{\sqrt{43}}{2}\right)}{6} + \left(\frac{2}{5} - \frac{\sqrt{43}}{2}\right)^{2}\right) - 2 \left(\frac{2}{5} - \frac{\sqrt{43}}{2}\right)\right) \leq 12$$

                 2               
     /      ____\      ____      
11   |2   \/ 43 |    \/ 43  <= 12
-- + |- - ------|  + ------      
10   \5     2   /      2         

but

                 2               
     /      ____\      ____      
11   |2   \/ 43 |    \/ 43  >= 12
-- + |- - ------|  + ------      
10   \5     2   /      2         

Then
$$x \leq \frac{1}{2} - \frac{\sqrt{43}}{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{1}{2} - \frac{\sqrt{43}}{2} \wedge x \leq \frac{1}{2} + \frac{\sqrt{43}}{2}$$

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