I'm stuck on this problem...what did i do wrong?

d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

cycles 4 meters per second

rate of .4 meters per second squared

vertical distance from top to bottom of hill is 25 meters

How long until goes down the hill?

I have...

(1/5)t^2+4t-25

From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?