(x-3)²-x(x-5)>13 inequation

Given the inequality:
$$- x \left(x - 5\right) + \left(x - 3\right)^{2} > 13$$
To solve this inequality, we must first solve the corresponding equation:
$$- x \left(x - 5\right) + \left(x - 3\right)^{2} = 13$$
Solve:
Given the equation:

(x-3)^2-x*(x-5) = 13

Expand expressions:

9 + x^2 - 6*x - x*(x - 5) = 13

9 + x^2 - 6*x - x^2 + 5*x = 13

Reducing, you get:

-4 - x = 0

Move free summands (without x)
from left part to right part, we given:
$$- x = 4$$
Divide both parts of the equation by -1

x = 4 / (-1)

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:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$- x \left(x - 5\right) + \left(x - 3\right)^{2} > 13$$
$$- \frac{\left(-41\right) \left(-5 + - \frac{41}{10}\right)}{10} + \left(- \frac{41}{10} - 3\right)^{2} > 13$$

131     
--- > 13
 10     

the solution of our inequality is:
$$x < -4$$

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