physics

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 30.0 m above its launch point
part A:What was the arrow's initial speed?part B:How long did it take for the arrow to first reach a height of 15.0 m above its launch point.

i did the part A y=v.t_g/2*t^2
v.=24.8m/s
but i cant calculate part b my answer is wrong pls help with part b thanks!

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  1. in Part A
    y = Vi t - 4.9 t^2
    30 = Vi(2)- 4.9(4)
    2 Vi = 49.6
    Vi = 24.8 agreed

    Part B
    15 = 24.8 t - 4.9 t^2
    4.9 t^2 - 24.8 t + 15 = 0

    t = [ 24.8 +/-sqrt(24.8^2-4*4.9*15) ]/9.81

    =[24.8+/- 17.9]/9.81
    use smaller t for on the way up
    .702 seconds

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