sqrt(4-2x)>=3 inequation

Given the inequality:
$$\sqrt{4 - 2 x} \geq 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{4 - 2 x} = 3$$
Solve:
Given the equation
$$\sqrt{4 - 2 x} = 3$$
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:
$$\left(\sqrt{4 - 2 x}\right)^{2} = 3^{2}$$
or
$$4 - 2 x = 9$$
Move free summands (without x)
from left part to right part, we given:
$$- 2 x = 5$$
Divide both parts of the equation by -2

x = 5 / (-2)

We get the answer: x = -5/2

$$x_{1} = - \frac{5}{2}$$
$$x_{1} = - \frac{5}{2}$$
This roots
$$x_{1} = - \frac{5}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{5}{2} + - \frac{1}{10}$$
=
$$- \frac{13}{5}$$
substitute to the expression
$$\sqrt{4 - 2 x} \geq 3$$
$$\sqrt{4 - \frac{\left(-13\right) 2}{5}} \geq 3$$

  _____     
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------- >= 3
   5        
     

the solution of our inequality is:
$$x \leq - \frac{5}{2}$$

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       x1