√(7-3x)>5 inequation

Given the inequality:
$$\sqrt{7 - 3 x} > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{7 - 3 x} = 5$$
Solve:
Given the equation
$$\sqrt{7 - 3 x} = 5$$
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:
$$\left(\sqrt{7 - 3 x}\right)^{2} = 5^{2}$$
or
$$7 - 3 x = 25$$
Move free summands (without x)
from left part to right part, we given:
$$- 3 x = 18$$
Divide both parts of the equation by -3

x = 18 / (-3)

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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-6 + - \frac{1}{10}$$
=
$$- \frac{61}{10}$$
substitute to the expression
$$\sqrt{7 - 3 x} > 5$$
$$\sqrt{7 - \frac{\left(-61\right) 3}{10}} > 5$$

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-------- > 5
   10       
    

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

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