HELP!! Chemistry Question!?

____________ A(g) ____ ⇌ ____ 2B(g) ____ Kc = 16.0
Initial: _____ 1.0 M __________ 0 M
Change: ____ -y M __________ +2y M
At eqm: __ (1.0 - y) M ________ 2y M

At eqm:
Kc = [B]² / [A]
16.0 = (2y)² / (1.0 - y)
16 - 16y = 4y²
4y² + 16y - 16 = 0
y² + 4y - 4 = 0
y = [-4 ± √(4² + 4*1*4)/2
y = 0.828 or y = -4.83 (rejected for y must be positive)

[B] at eqm = 2y M = 2 × 0.828 M = 1.66 M ≈ 1.7 M (to 2 sig. fig)

Correct answer:
INITIAL [A] = 1.0 M, NOT equilibrium concentration.

Kc = [B]^2 / [A] = 16.0
At equilibrium, let [B] = 2x. Then, at equilibrium, [A] = 1-x. Then,

Kc = 16 = (2x)2^2 / (1-x)
16 - 16x = 4x^2
4x^2 + 16x - 16 = 0

Using the quadratic formula, the valid root is x = 0.828. So,
[B] = 2(0.828) = 1.656 M
[A] = 1 - 0.828 = 0.272 M

I would round answers to 2 significant figures.

Kc=[B]^2 / [A] <=> 16 = [B]^2 / 1 <=> [B] = 4 M. [NOTE: Wrong solution, read further.]
(Answer to comment:) I'm sure this is the right answer. Both A and B are gases, therefore both B and A concentrations are in the Kc fraction. I don't know what may be wrong.

(Answer to comment 2:) I'm sure your computer is false. That's the right answer for sure. What are you doing for the calculations in your computer?
Unless one of those substances isn't a gas, the answer is correct.

(Answer to comment 3:) I'm not sure in which way you're using your computer to check the answer. Please, make sure you wrote the exercise correctly (the concentrations, the states and the equillibrium). If everything is OK, then I would say don't trust the computer.

(Answer to comment 4:) OH WAIT It's INITIAL CONCENTRATION. You're right. Wait, I'll fix it.

I GOT IT. You subtract x from A concentration and 2x from B. Kc = 4x^2/(1-x) You get x = (-4 + sqrt (32) )/2 which is B concentration.

Well, I guess I'm late and lost best answer unfortunately... Stupid "initial" concentration...

Kc = [B]^2 / [A]
Hence
[B]^2 = Kc [A]
[B] = sqrt{ Kc [ A]}
[B] = sqrt{ 16.00 x 1.0
[B] = sqrt16
[B] = 4 M