Please explain to work this:

A rocket is launched from atop a 105-foot cliff with an initial velocity of 156ft/s. the height of the rocket above the ground at time t is given by

h= -16t^2 + 156t + 105. When will the

rocket hit the ground after it is launched? Round to the nearest second.

naturally, when h = 0.

-16t^2 + 156t + 105 = 0
t = (39 ± √1941)/8
t = 10 sec

???????????????????????????????????????Your confusing me! STOP CONFUSING ME!!!!!

h=0, so you set this up in the quadratic formula, which then you follow all the rules and your left with a positive and negative answer. time cant be negative, so the correct answer should be about 10.38 seconds which rounds off to 10.4

I agree with anonymous

To find when the rocket hits the ground, we need to determine the value of t when the height (h) is equal to 0.

In the given equation, the height of the rocket above the ground (h) is given by:

h = -16t^2 + 156t + 105

Setting h to zero and solving for t:
0 = -16t^2 + 156t + 105

This equation is a quadratic equation, so we can solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -16, b = 156, and c = 105.

Now we can plug these values into the quadratic formula:

t = (-156 ± √(156^2 - 4*(-16)*105)) / (2*(-16))

Simplifying further:

t = (-156 ± √(24336 + 6720)) / (-32)
t = (-156 ± √(31056)) / (-32)
t = (-156 ± 176) / (-32)

Now we can calculate the two possible values for t:

t1 = (-156 + 176) / (-32) = 20 / (-32) = -0.625
t2 = (-156 - 176) / (-32) = -332 / (-32) = 10.375

Since time cannot be negative in this context, we can ignore the negative value of t (-0.625).

Therefore, the rocket will hit the ground approximately 10 seconds after it is launched.