Finding the time when two stones cross paths

1. The problem statement, all variables and given/known data
Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.69 m. The stones are thrown with the same speed of 9.97 m/s. After what time do the stones cross paths?

2. Relevant equations
d = vt + 0.5at2

3. The attempt at a solution
distance travelled by both rock A and rock B combined is 6.69
DA + DB = 6.69

let the upwards direction be positive

DA: (top of cliff)

d = vit + 0.5at2
= (-9.97 m/s[up])t + 0.5(-9.8 m/s2[up])t2
= -9.97t – 4.9t2

DB: (base of cliff)
d = vit + 0.5at2
d = (9.97 m/s [up])t + 0.5(-9.8m/s2[up])t2
d = 9.97t – 4.9t2

DA + DB = 6.69
-9.97t – 4.9t2 + 9.97t – 4.9t2 = 6.69
-9.8t2 = 6.69 getting negative time…?
t = 0.82 s

the correct answer is 0.335 seconds. I think I am messing up my directions and signs for my vectors… any help?

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