What is the equilibrium partial pressure of H₂?

Question

An empty steel container is filled with 0.0400 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H₂? 2 HF (g) ⇌ H₂ (g) + F₂ (g)

Answer 1

2HF(g) <====> H₂(g) + F₂(g)

0.0400................0................0...........Initial

-2x...................+x...............+x...........Change

0.04-2x.............x.................x............Equilibrium

Kp = 2.76 = (P_H2)(P_F2) / (P_HF)²

2.76 = (x)(x) / (0.04-2x)²

Solve for x

I got x = 0.0292 atm = equilibrium partial pressure of H₂, but please do check my math.