sqrt(2x-3)<3 inequation

Given the inequality:
$$\sqrt{2 x - 3} < 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2 x - 3} = 3$$
Solve:
Given the equation
$$\sqrt{2 x - 3} = 3$$
Because equation degree is equal to = 1/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2-th degree:
We get:
$$\left(\sqrt{2 x - 3}\right)^{2} = 3^{2}$$
or
$$2 x - 3 = 9$$
Move free summands (without x)
from left part to right part, we given:
$$2 x = 12$$
Divide both parts of the equation by 2

x = 12 / (2)

We get the answer: x = 6

$$x_{1} = 6$$
$$x_{1} = 6$$
This roots
$$x_{1} = 6$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 6$$
=
$$\frac{59}{10}$$
substitute to the expression
$$\sqrt{2 x - 3} < 3$$
$$\sqrt{-3 + \frac{2 \cdot 59}{10}} < 3$$

    ____    
2*\/ 55     
-------- < 3
   5        
    

the solution of our inequality is:
$$x < 6$$

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       x1