Algebra
posted by Selena .
A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = 16t2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?

16 t^2 20 t  6 =  h in obsolete English units
when does h = 0?
t = [ 20 +/ sqrt (400 + 384) ] /32
t = [ 20 +/ 28 ] /32
t = 48/32 = 1.5 seconds (ignore the negative t)
for vertex complete the square assuming no calculus may be used
16 t^2 20 t =  h + 6
t^2  (5/4) t = h/16 + 3/16
t^2  (5/4)t + 25/64 = 4h/64 + 37/64
(t5/8)^2 = (4/64) (h  148 )
max height = 148 at t = 5/8 sec 
Find the vertex of the parabola by completing the square:
h(t)=16t²+20t+6
=16(t²2(5/8)t+25/64)+6+25/4
which means that the vertex is at
t=5/8
and it will hit the ground at h(t)=0, or
t=3/2 s (rejecting negative root).
Maximum is at the vertex, or
h(5/8)=49/4 ft. 
thanx both of yall XD
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