9x^2-12x+4≤0 inequation

Given the inequality:
$$9 x^{2} - 12 x + 4 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$9 x^{2} - 12 x + 4 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 9$$
$$b = -12$$
$$c = 4$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 9 \cdot 4 \cdot 4 + \left(-12\right)^{2} = 0$$
Because D = 0, then the equation has one root.

x = -b/2a = --12/2/(9)

$$x_{1} = \frac{2}{3}$$
$$x_{1} = \frac{2}{3}$$
$$x_{1} = \frac{2}{3}$$
This roots
$$x_{1} = \frac{2}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{2}{3}$$
=
$$\frac{17}{30}$$
substitute to the expression
$$9 x^{2} - 12 x + 4 \leq 0$$
$$- \frac{12 \cdot 17}{30} + 9 \left(\frac{17}{30}\right)^{2} + 4 \leq 0$$

9/100 <= 0

but

9/100 >= 0

Then
$$x \leq \frac{2}{3}$$
no execute
the solution of our inequality is:
$$x \geq \frac{2}{3}$$

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