SOLUTION: The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest       Log On


   

Question 289365: The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest tenth) is the object 11 feet in the air?
Could you help to solve this equation?
Thank you.
Cynthia
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest tenth) is the object 11 feet in the air?
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h is the height so make it 11 and solve for "t":
t^2-7t+3 = 11
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t^2 -7t -8 = 0
(t-8)(t+1) = 0
Positive solution:
t = 8 seconds (time at which the object is at a height of 11 ft.
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Cheers,
Stan H.