Anyone good at physics?

Question

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. (a) What was the arrow's initial speed? (b) How long did it take for the arrow to first reach a heightof 15.0 above its launch point?

Answer 1

Let s(zero) = 0
(a)
s = s(zero) + v(zero)t +(1/2)gt^2
v(zero)t = s - (1/2)gt^2
v(zero) = (s - (1/2)gt^2)/t
v(zero) = (30 - (1/2)(9.8)2^2)/2
v(zero) = 5.2m/s

(b):
s = v(zero) t + (1/2)gt^2
(1/2)gt^2 + v(zero)t -s = 0
(1/2)(9.8)t^2 + 5.2t - 15 = 0
4.9t^2 + 5.2t - 15 = 0
t = (-B +/- √(B^2 - 4AC))/2A
t = {-5.2 +/- √[(5.2)^2 - 4(4.5)(-15)]}/2(4.9)
t = (-5.2 +/- 17.91756)/9.8
Only the positive answer makes sense:
t = 12.7176/9.8 = 1.298s