Answer:
43.95mph
Step-by-step explanation:
let x=returning speed
x-30=foward speed
Travel time=distance/speed
180/(x-30)-180/x=3
LCD: x(x-30)
180x-180(x-30)=3x(x-3)
180x-180x+5400=3x^2-9x
3x^2-9x-5400=0
Use quadratic formula to solve
(-b+-√b^2-4ac) / 2a.
a = 3, b= -9, c = - 5400
x=[-(-9)±sqrt((-9)^2-4*3*-5400)]/2*3
x = 9±sqrt(81+64800)]/2*3
x = 9±sqrt(64881)]/6
x = 9±254.717/6
x = 263.717/6 or -254.717/6
x = 43.95 or - 42.45( will be rejected because it is less than 0)
x = 43.95
The returning speed will be 43.95mph.
