0,5^(2x-1)+3×(0,5)^x-2>=0 inequation

Given the inequality:
$$\left(\left(\frac{1}{2}\right)^{2 x - 1} + 3 \left(\frac{1}{2}\right)^{x}\right) - 2 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\left(\frac{1}{2}\right)^{2 x - 1} + 3 \left(\frac{1}{2}\right)^{x}\right) - 2 = 0$$
Solve:
Given the equation:
$$\left(\left(\frac{1}{2}\right)^{2 x - 1} + 3 \left(\frac{1}{2}\right)^{x}\right) - 2 = 0$$
or
$$\left(\left(\frac{1}{2}\right)^{2 x - 1} + 3 \left(\frac{1}{2}\right)^{x}\right) - 2 = 0$$
Do replacement
$$v = \left(\frac{1}{4}\right)^{x}$$
we get
$$2^{1 - 2 x} - 2 + 3 \cdot 2^{- x} = 0$$
or
$$2^{1 - 2 x} - 2 + 3 \cdot 2^{- x} = 0$$
do backward replacement
$$\left(\frac{1}{4}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
$$x_{1} = 0.999999999999577$$
$$x_{2} = 1$$
$$x_{1} = 0.999999999999577$$
$$x_{2} = 1$$
This roots
$$x_{1} = 0.999999999999577$$
$$x_{2} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 0.999999999999577$$
=
$$0.899999999999577$$
substitute to the expression
$$\left(\left(\frac{1}{2}\right)^{2 x - 1} + 3 \left(\frac{1}{2}\right)^{x}\right) - 2 \geq 0$$
$$-2 + \left(\left(\frac{1}{2}\right)^{-1 + 0.899999999999577 \cdot 2} + \frac{3}{2^{0.899999999999577}}\right) \geq 0$$

0.182009371303766 >= 0

one of the solutions of our inequality is:
$$x \leq 0.999999999999577$$

 _____           _____          
      \         /
-------•-------•-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 0.999999999999577$$
$$x \geq 1$$