To solve the inequality A ≥ 0.5, we need to find the values of a that satisfy this condition.

Given the equation A = √(1-4a+4a^2) - 2a, we can rewrite it as:

A + 2a = √(1-4a+4a^2)

Squaring both sides, we get:

(A + 2a)^2 = 1-4a+4a^2

Expanding the left side, we have:

A^2 + 4aA + 4a^2 = 1-4a+4a^2

Rearranging the terms, we get:

A^2 + 4aA + 4a^2 - 1 + 4a - 4a^2 = 0

Simplifying further, we have:

A^2 + 4aA - 1 + 4a = 0

Now, we can solve this quadratic equation for A in terms of a using the quadratic formula:

A = (-4a ± √(4a^2 - 4(1)(-1 + 4a))) / 2

Simplifying further, we have:

A = (-4a ± √(4a^2 + 4 + 16a)) / 2

A = (-4a ± √(4(a^2 + 1 + 4a))) / 2

A = (-4a ± 2√(a^2 + 1 + 4a)) / 2

A = -2a ± √(a^2 + 1 + 4a)

Now, we can substitute this expression for A back into the inequality A ≥ 0.5:

-2a ± √(a^2 + 1 + 4a) ≥ 0.5

We can solve this inequality by considering the two cases separately:

Case 1: -2a + √(a^2 + 1 + 4a) ≥ 0.5

Simplifying further, we have:

√(a^2 + 1 + 4a) ≥ 0.5 + 2a

Squaring both sides, we get:

a^2 + 1 + 4a ≥ 0.25 + 2a + 4a^2

Rearranging the terms, we have:

3a^2 + 2a - 0.75 ≥ 0

Using the quadratic formula, we can find the values of a that satisfy this inequality.

Case 2: -2a - √(a^2 + 1 + 4a) ≥ 0.5

Simplifying further, we have:

-√(a^2 + 1 + 4a) ≥ 0.5 + 2a

Squaring both sides, we get:

a^2 + 1 + 4a ≥ 0.25 + 2a + 4a^2

Rearranging the terms, we have:

3a^2 + 2a - 0.75 ≤ 0

Using the quadratic formula, we can find the values of a that satisfy this inequality.

In summary, to find the values of a that satisfy the inequality A ≥ 0.5, we need to solve the quadratic inequality 3a^2 + 2a - 0.75 ≥ 0. The solutions to this inequality will give us the range of values for a that satisfy the given condition.