(4^x-2^x+3)+28(4^x-2^x+3)+192>=0 inequation

Given the inequality:
$$- 2^{x} + 4^{x} + 28 \left(- 2^{x} + 4^{x} + 3\right) + 3 + 192 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- 2^{x} + 4^{x} + 28 \left(- 2^{x} + 4^{x} + 3\right) + 3 + 192 = 0$$
Solve:
Given the equation:
$$- 2^{x} + 4^{x} + 28 \left(- 2^{x} + 4^{x} + 3\right) + 3 + 192 = 0$$
or
$$\left(- 2^{x} + 4^{x} + 28 \left(- 2^{x} + 4^{x} + 3\right) + 3 + 192\right) + 0 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$29 v^{2} - 29 v + 279 = 0$$
or
$$29 v^{2} - 29 v + 279 = 0$$
This equation is of the form
$$a\ v^2 + b\ v + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 29$$
$$b = -29$$
$$c = 279$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 29 \cdot 4 \cdot 279 + \left(-29\right)^{2} = -31523$$
Because D<0, then the equation
has no real roots,
but complex roots is exists.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = \frac{1}{2} + \frac{\sqrt{31523} i}{58}$$
Simplify
$$v_{2} = \frac{1}{2} - \frac{\sqrt{31523} i}{58}$$
Simplify
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{31523} i}{58}$$
$$x_{2} = \frac{1}{2} - \frac{\sqrt{31523} i}{58}$$
Exclude the complex solutions:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example
$$x_0 = 0$$
$$- 2^{0} + 4^{0} + 3 + 28 \left(- 2^{0} + 4^{0} + 3\right) + 192 \geq 0$$

279 >= 0

so the inequality is always executed