Solve 16= 2(150) (S+42)/S^2

Question

Bungee Jumping Originating on Pentecost Island in the Pacific, the practice of a person jumping from a high place harnessed to a flexible attachment was introduced to Western culture in 1979 by the Oxford University Dangerous Sport Club. One important parameter to know before attempting a bungee jump is the amount the cord will stretch at the bottom of the fall. The stiffness of the cord is related to the amount of stretch by the equation K = 2W (S + L)/S^2 where W =weight of the jumper (pounds) K = cord’s stiffness (pounds per foot) L = free length of the cord (feet) S = stretch (feet) A 150-pound person plans to jump off a ledge attached to a cord of length 42 feet. If the stiffness of the cord is no less than 16 pounds per foot, how much will the cord stretch?

Answer 1

2*150*(x+42)/ x^2=16... 16x^2=300(x+42)... 16x^2 -300x -12600=0... or 4x^2 -75x -3150=0... quadratik equashun giv roots: x=38.9620009... (ignore negativ 20.2120009)