Algebra Distance Problem Help?
The speed of a moving walkway is 2.6 feet per second. It takes Kim a total of 50 seconds to travel 65 feet with the movement of the walkway, and back, with the movement against the walkway. What is Kim's normal walking speed???
I understand the basics of this question, and the t=s/v However, something goes wrong for me when I get to the quadratic formula portion of the equation. Can someone please give a detailed answer? THANKS!
Update:Awesome! I appreciate the help - thanks so much! 
Comments
n = normal walking speed
t = time walking with walkway
50 - t = time against
65 = (n + 2.6) • t
65 = (n - 2.6) (50 - t)
From the first, 65/(n + 2.6) =t
So that makes the second be 65 = (n - 2.6)[50 - 65/(n + 2.6)]
FOIL it out
65 = 50n - 65n/(n + 2.6) - 130 + 169/(n + 2.6)
Multiply by (n + 2.6)
65(n + 2.6) = 50n(n + 2.6) - 65n - 130(n + 2.6) + 169
65n + 169 = 50n^2 + 130n - 65n - 130n - 338 + 169
0 = 50n^2 + 130n - 65n - 130n - 338 + 169 - 65n - 169
0 = 50n^2 - 130n - 338
0 = 25n^2 - 65n - 169
so by the QF, n = 65 + √(4225 - 4(25)(-169)) over 2(25)
n = (65 + 145.3444)÷50
n is about 4.21 ft/sec