Can anyone help me with this algebra question?

Question

Brad tossed a baseball in the air from a height of 6 feet with an initial upward velocity of 14 feet per second. Enrique caught the ball on its way down at a point 4 feet above the ground. How long was the ball in the air before Enrique caught it? Use the equation h=-16t^2+vt+s where 'h' is height in feet, 't' is time in seconds, 'v' is the initial upward velocity in feet per second, and 's' is the initial height in feet. (let h=0)

Answer 1

h(t) = -16t² + 14t + 6

So you want h(t) to be equal to 4:

4 = -16t² + 14t + 6
0 = -16t² + 14t + 2

Divide through by -2:

0 = 8t² - 7t - 1
0 = (8t + 1)(t - 1)

So the ball would be 4 feet above the ground at t = 1 second (throwing out the -1/8 answer for the same reason as the last problem you asked like this one).

Answer 2

The height of the ball is expressed by the formula:

h = -16t² + vt + s

You are given

v = 14 ft/sec
s = 6 ft

So

h = -16t² + vt + s = -16t² + 14t + 6

The ball is caught at h = 4 ft. So we have

h = -16t² + 14t + 6 = 4
-16t² + 14t + 2 = 0
8t² - 7t - 1 = 0
(8t - 1)(t + 1) = 0
t = -1/8 and 1

But time must be positive so t = 1 second.

Your last statement "(let h=0)" contradicts ealier statements about the initial and final height of the ball, so I ignored it.