2*sqrt(x)-3-sqrt(x)+2>=1 inequation

Given the inequality:
$$\left(- \sqrt{x} + \left(2 \sqrt{x} - 3\right)\right) + 2 \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- \sqrt{x} + \left(2 \sqrt{x} - 3\right)\right) + 2 = 1$$
Solve:
Given the equation
$$\left(- \sqrt{x} + \left(2 \sqrt{x} - 3\right)\right) + 2 = 1$$
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:
$$\left(\sqrt{x}\right)^{2} = 2^{2}$$
or
$$x = 4$$
We get the answer: x = 4

$$x_{1} = 4$$
$$x_{1} = 4$$
This roots
$$x_{1} = 4$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 4$$
=
$$\frac{39}{10}$$
substitute to the expression
$$\left(- \sqrt{x} + \left(2 \sqrt{x} - 3\right)\right) + 2 \geq 1$$
$$\left(- \sqrt{\frac{39}{10}} + \left(-3 + 2 \sqrt{\frac{39}{10}}\right)\right) + 2 \geq 1$$

       _____     
     \/ 390      
-1 + ------- >= 1
        10       
     

but

       _____    
     \/ 390     
-1 + ------- < 1
        10      
    

Then
$$x \leq 4$$
no execute
the solution of our inequality is:
$$x \geq 4$$

         _____  
        /
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       x1