Concentration of B at equilibrium? ?

               A(aq)   ⇌   2B(aq)   Kc = 5.85 × 10⁻⁶
Initial:     3.10 M          0 M
Change:  -y M          +2y M
Eqm:   (3.10 - y) M     2y M

At equilibrium:
Kc = [B]² / A
5.85 × 10⁻⁶ = (2y)² / (3.10 - y)
(5.85 × 10⁻⁶)(3.10 - y) = (2y)²
4y² + (5.85 × 10⁻⁶)y - (1.8135 × 10⁻⁵) = 0
y = {-(5.85 × 10⁻⁶) + √[(5.85 × 10⁻⁶)² + 4×4×(1.8135 × 10⁻⁵)]} / (2*4)
y = 2.13 × 10⁻³
2y = 4.26 × 10⁻³

[B] at equilibrium = 4.26 × 10⁻³ M