Anyone who's good in math! HELP!

If (x-4)^4 =49 and x>0, what is the value of x^2?

(x-4) = sqrt(sqrt(49)) = sqrt(7)
=> x = 4+sqrt(7)
=> x² = 16+8sqrt(7)+7 = 23+8sqrt(7)
is one solution
x = 4-sqrt(7) is another solution
=> x² = 23-8sqrt(7)
the other two solutions are complex

(x - 4)^4 = 49 and x > 0 suggests to me that (x - 4)² = 7
x² - 8x + 16 = 7

x² - 8x + 9 = 0
Use the quadratic formula or complete the square to solve, and take the positive value of x.

x - 4 = ±49^(1/4)
x - 4 = ±(7^2)^(1/4)
x - 4 = ±7^(1/2) = ±√7
x = 4 ±√7
x^2 = (4 ±√7)^2
= 16 ± 8√7 + 7
= 23 ± 8√7
≈ 1.83 or 44.16

if (x-4)^4=49

x-4= 4th root(49)

x=4th root(49) +4 this is gonna be ugly

x=6.646
x^2=44.17

It's approximately 44.16