An object is launched at 9.8 meters per second from a 73.5-meter tall platform. The object's height s (in meters) after t seconds is given by the equation s(t)= -4.9t(-4.9t) - 9.8t = 73.5. When does the object strike the ground?

I can't figure out how to factor it fully. I got to -4.9(t^2 +2t +15) = 0

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  1. it was evidently launched straight up in which case:
    h(t) = 73.5 + 9.8 t - (1/2)g t^2
    g is about 9.81 m/s^2 so

    h(t) = 73.5 + 9.8 t - 4.9 t^2

    at ground, h(t) = 0
    0 = 4.9 t^2 - 9.8 t - 73.5
    and yes you can take 4.9 out approximately
    0 = t^2 - 2 t -15
    (t-5)(t+3) = 0
    t = 5 seconds

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  2. factor out -4.9
    then think factors of -15 that will give you a positive 2. 5, -3 then set it up to zero because the height is at zero since it is on the ground.
    when solving you would get
    t+5=0 which means t= -5 secs
    t-3=0 which means t= 3 sec
    Since time cannot be negative
    your answer would be t= 3 secs

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