(4^x-2^x+2) ^2-28(4^x-2^x+2)-128≥0

Let's make a substitution to simplify the expression. Let y = 4^x - 2^x + 2.

Now the inequality becomes y^2 - 28y - 128 ≥ 0.

To solve this quadratic inequality, we first find the roots of the quadratic equation y^2 - 28y - 128 = 0 by factoring or using the quadratic formula:

y^2 - 28y - 128 = 0 (y - 32)(y + 4) = 0 y = 32 or y = -4

Now we have three intervals to test: (-∞, -4), (-4, 32), and (32, ∞).

For y < -4: y^2 - 28y - 128 > 0

For -4 < y < 32: y^2 - 28y - 128 < 0

For y > 32: y^2 - 28y - 128 > 0

Therefore, the solution to the inequality (4^x - 2^x + 2) ^2 - 28(4^x - 2^x + 2) - 128 ≥ 0 is y ≤ -4 or y ≥ 32, which translates back to 4^x - 2^x + 2 ≤ -4 or 4^x - 2^x + 2 ≥ 32.