CHEMISTRY HELP PLEASE??Calculate the equilibrium partial pressures of both species when the initial pressure of NO2 is 0.050 atm.?

              N₂O₄(g)    ⇌    2NO2       Kp = 0.25
Initial:       0 atm         0.05 atm
Change:  +y atm         -2y atm
Eqm:        y atm    (0.050 - 2y) atm

At equilibrium:
Kp = 𝑃NO₂² / 𝑃N₂O₄
0.25 = (0.050 - 2y)² / y
0.25y = 0.0025 - 0.2y + 4y²
4y² - 0.45y + 0.0025 = 0
y = [0.45 ± √(0.45² - 4*4*0.0025)] / (2*4)
y = 0.00586  or  y = 0.107 (rejected for 0.05 - 2y < 0)

Equilibrium pressure of N2O4 = 0.00586 atm ≈ 0.0059 atm
Equilibrium pressure of NO2 = (0.050 - 2*0.00586) atm ≈ 0.038 atm
(All answers are to 2 sig. fig.)

I won't solve your specific problem, but I do have some related examples here:

https://www.chemteam.info/Equilibrium/Calc-equib-from-init-cond.html

Here's one comment: If the (N2O4) at equilibrium is x, the the (NO2) that was used up is 2x.

Here's a second: the equilibrium concentration of NO2 will be 0.050 - 2x