\root(3)(3x-8)<-2 inequation

Given the inequality:
$$\sqrt{3} \left(3 x - 8\right) < -2$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{3} \left(3 x - 8\right) = -2$$
Solve:
Given the equation:

sqrt(3)*(3*x-8) = -2

Expand expressions:

-8*sqrt(3) + 3*x*sqrt(3) = -2

Reducing, you get:

2 - 8*sqrt(3) + 3*x*sqrt(3) = 0

Expand brackets in the left part

2 - 8*sqrt3 + 3*x*sqrt3 = 0

Move free summands (without x)
from left part to right part, we given:
$$3 \sqrt{3} x - 8 \sqrt{3} = -2$$
Divide both parts of the equation by (-8*sqrt(3) + 3*x*sqrt(3))/x

x = -2 / ((-8*sqrt(3) + 3*x*sqrt(3))/x)

We get the answer: x = 8/3 - 2*sqrt(3)/9
$$x_{1} = \frac{8}{3} - \frac{2 \sqrt{3}}{9}$$
$$x_{1} = \frac{8}{3} - \frac{2 \sqrt{3}}{9}$$
This roots
$$x_{1} = \frac{8}{3} - \frac{2 \sqrt{3}}{9}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(\frac{8}{3} - \frac{2 \sqrt{3}}{9}\right)$$
=
$$\frac{77}{30} - \frac{2 \sqrt{3}}{9}$$
substitute to the expression
$$\sqrt{3} \left(3 x - 8\right) < -2$$
$$\sqrt{3} \left(-8 + 3 \left(\frac{77}{30} - \frac{2 \sqrt{3}}{9}\right)\right) < -2$$

      /           ___\     
  ___ |  3    2*\/ 3 |     
\/ 3 *|- -- - -------| < -2
      \  10      3   /     
     

the solution of our inequality is:
$$x < \frac{8}{3} - \frac{2 \sqrt{3}}{9}$$

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