Question

The height of a projectile fired upward is given by the formula s = v0t – 16t2, where s is the height, v0 is the initial velocity and t is the time. Find the times at which a projectile with an initial velocity of 128 ft/s will be 144 ft above the ground. Round to the nearest hundredth of a second.
t = sec (smaller value)
t = sec (larger value)

Answer 1

So you are solving:

144 = 128t - 16t^2

16t^2 - 128t + 144 = 0
t^2 - 8t + 9 = 0

I will complete the square, (in this case faster than the formula)
t^2 - 8t + 16 = -9+16
(t-4)^2 = 7
t-4 = ±√7
t = 4 ± √7
I get appr 6.65 and 1.35

Answer 2

h = hi + Vi t - (1/2) g t^2
in ancient English units this is

h = Hi + Vi t - (1/2) 32 t^2

If
Hi = 0
Vi = 128 ft/s
then
h = 128 t - 16 t^2
for h = 144
144 = 128 t - 16 t^2

16 t^2 - 128 t + 144 = 0

t^2 - 8 t + 9 = 0

t = [ 8 +/- sqrt(64 -36) ]/2

t = 4 +/- sqrt 7

t = 1.35
t = 6.65