(0,2)^2-3x>5 inequation

Given the inequality:
$$- 3 x + \left(\frac{1}{5}\right)^{2} > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$- 3 x + \left(\frac{1}{5}\right)^{2} = 5$$
Solve:
Given the linear equation:

((1/5))^2-3*x = 5

Expand brackets in the left part

1/5)^2-3*x = 5

Move free summands (without x)
from left part to right part, we given:
$$- 3 x = \frac{124}{25}$$
Divide both parts of the equation by -3

x = 124/25 / (-3)

$$x_{1} = - \frac{124}{75}$$
$$x_{1} = - \frac{124}{75}$$
This roots
$$x_{1} = - \frac{124}{75}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{124}{75} + - \frac{1}{10}$$
=
$$- \frac{263}{150}$$
substitute to the expression
$$- 3 x + \left(\frac{1}{5}\right)^{2} > 5$$
$$\left(\frac{1}{5}\right)^{2} - \frac{\left(-263\right) 3}{150} > 5$$

53    
-- > 5
10    

the solution of our inequality is:
$$x < - \frac{124}{75}$$

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