9x^2-12x-4≤0 inequation

Given the inequality:
$$\left(9 x^{2} - 12 x\right) - 4 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(9 x^{2} - 12 x\right) - 4 = 0$$
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
$$a = 9$$
$$b = -12$$
$$c = -4$$
, then

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

(-12)^2 - 4 * (9) * (-4) = 288

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

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

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

or
$$x_{1} = \frac{2}{3} + \frac{2 \sqrt{2}}{3}$$
$$x_{2} = \frac{2}{3} - \frac{2 \sqrt{2}}{3}$$
$$x_{1} = \frac{2}{3} + \frac{2 \sqrt{2}}{3}$$
$$x_{2} = \frac{2}{3} - \frac{2 \sqrt{2}}{3}$$
$$x_{1} = \frac{2}{3} + \frac{2 \sqrt{2}}{3}$$
$$x_{2} = \frac{2}{3} - \frac{2 \sqrt{2}}{3}$$
This roots
$$x_{2} = \frac{2}{3} - \frac{2 \sqrt{2}}{3}$$
$$x_{1} = \frac{2}{3} + \frac{2 \sqrt{2}}{3}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(\frac{2}{3} - \frac{2 \sqrt{2}}{3}\right) + - \frac{1}{10}$$
=
$$\frac{17}{30} - \frac{2 \sqrt{2}}{3}$$
substitute to the expression
$$\left(9 x^{2} - 12 x\right) - 4 \leq 0$$
$$-4 + \left(9 \left(\frac{17}{30} - \frac{2 \sqrt{2}}{3}\right)^{2} - 12 \left(\frac{17}{30} - \frac{2 \sqrt{2}}{3}\right)\right) \leq 0$$

                                 2     
                   /         ___\      
  54       ___     |17   2*\/ 2 |  <= 0
- -- + 8*\/ 2  + 9*|-- - -------|      
  5                \30      3   /      

but

                                 2     
                   /         ___\      
  54       ___     |17   2*\/ 2 |  >= 0
- -- + 8*\/ 2  + 9*|-- - -------|      
  5                \30      3   /      

Then
$$x \leq \frac{2}{3} - \frac{2 \sqrt{2}}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{2}{3} - \frac{2 \sqrt{2}}{3} \wedge x \leq \frac{2}{3} + \frac{2 \sqrt{2}}{3}$$

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