A snowball was thrown from a cliff. The height above ground of the snowball is modeled by the function h(t) = -16t2 + 48t + 28, where h is height in feet, and t is time in seconds. How long was the snowball in the air?

Did everything got (4t+7)(t+1)
then did 4t+7=0 i got t=-1.75 then i did the x+1=0 i got -1, its asking for how long it was in the air(seconds) what did i do wrong or how do i find the answer.

Snowball was in the air as long as h > 0

let's see when h = 0
0 = -16t^2 + 48t + 28
divide by -4
4t^2 - 12t - 7 = 0
(2t + 1)(2t - 7) = 0
so t = -1/2 or t = 7/2
but t ≥ 0
so the ball was in the air for 7/2 or 3.5 seconds

If you plug it in to a calculator you get 3.5

You didn't factor correctly
(4t+7)(t+1)=4t^2+11t+7

-16t^2+48t+28
-16t^2-8t+56t+28
(16t+8)(-t+3.5)
t=-1/2 t=3.5

To find how long the snowball was in the air, we need to determine the values of t when the height is 0. In other words, we are looking for the time at which the height function h(t) = -16t^2 + 48t + 28 equals zero.

To solve the equation -16t^2 + 48t + 28 = 0, we can apply the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation.

For the given equation, a = -16, b = 48, and c = 28. Plugging these values into the quadratic formula, we get:

t = (-48 ± √(48^2 - 4(-16)(28))) / (2(-16))
t = (-48 ± √(2304 + 1792)) / (-32)
t = (-48 ± √(4096)) / (-32)
t = (-48 ± 64) / (-32)

Now, we have two possible solutions:

1. t = (-48 + 64) / (-32) = 16 / (-32) = -0.5
2. t = (-48 - 64) / (-32) = -112 / (-32) = 3.5

Since time cannot be negative in this context, the snowball was in the air for 3.5 seconds.