√x+2>2 inequation

Given the inequality:
$$\sqrt{x} + 2 > 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{x} + 2 = 2$$
Solve:
Given the equation
$$\sqrt{x} + 2 = 2$$
so
$$x = 0$$
We get the answer: x = 0
$$x_{1} = 0$$
$$x_{1} = 0$$
This roots
$$x_{1} = 0$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10}$$
=
$$- \frac{1}{10}$$
substitute to the expression
$$\sqrt{x} + 2 > 2$$
$$2 + \sqrt{- \frac{1}{10}} > 2$$

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    I*\/ 10     
2 + -------- > 2
       10       
    

Then
$$x < 0$$
no execute
the solution of our inequality is:
$$x > 0$$

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       x1