let

x = spped of the wind

v = speed of the plane in still air

d = 715 miles the distance

t1 = 2 hours 10 min = 2 10/60 hours = 13/6 hours the time of the trip

t2 = 1 hour 50 min = 1 50/60 = 11/6 hours the time of the return trip

since t2 < t1 in the return trip the plane is flying in the direction of the wind while at the beginning it was flying against the speed of the wind:

the speed of the plane against the wind is v - x = 715 / (13/6) miles per hour

the speed of the plane in the return trip is v + x = 715 / (11/6) miles per hour

thus we can write

v - x = 715/(13/6)

v + x = 715/(11/6)

by solving the system of equations

v - x = 330

v + x = 390

we find

v = 360 mph

x = 30 mph

the speed of the plane in still air is 360 mph.

the speed of the wind is 30 mph.