Projectile MATH Problem

Question

A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 109 feet per second from an initial height of 112 feet off the ground, then the height of the projectile, , in feet,  seconds after it's shot is given by the equation:

Find the two points in time when the object is 159 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).

Answer: The object is 159 feet off the ground at the following times...?????

Answer 1

Equation for h(t) = 112 + 109t - 16t². So to get this moments we have equation:

159 = 112 + 109t - 16t²; 16t² - 109t + 47 = 0, t = (109 ± √(109² - 64·47)/32

t₁ = 0.46 s, t₂ = 6.35 s,