16^2-8x+1>0 inequation

Given the inequality:
$$\left(256 - 8 x\right) + 1 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(256 - 8 x\right) + 1 = 0$$
Solve:
Given the linear equation:

16^2-8*x+1 = 0

Move free summands (without x)
from left part to right part, we given:
$$- 8 x = -257$$
Divide both parts of the equation by -8

x = -257 / (-8)

$$x_{1} = \frac{257}{8}$$
$$x_{1} = \frac{257}{8}$$
This roots
$$x_{1} = \frac{257}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{257}{8}$$
=
$$\frac{1281}{40}$$
substitute to the expression
$$\left(256 - 8 x\right) + 1 > 0$$
$$\left(256 - \frac{8 \cdot 1281}{40}\right) + 1 > 0$$

4/5 > 0

the solution of our inequality is:
$$x < \frac{257}{8}$$

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