Find position from acceleration, antiderivatives

We're given:
a(t)= t^2 -7t + 3
s(0)= 0 and s(1)= 20

so for the velocity formula I got:
1/3 t^3 - 7/2 t^2 + 3t + C

and for the speed:
1/12 t^4 - 7/6 t^3 + 3/2 t^2 + C + D

but since s(0)= 0 then c would be 0 and then we can sub in s(1)= 20 to find D. I did that and got D= 235/12
what's wrong?!

Although it didn't effect your work, s(t) represents position as you said in the question, not speed.
The only mistake you have is that since 'C' is a constant, when you take the antiderivative to go from velocity to position, the 'C' becomes 'Ct' just like how the 3 became 3t when you got velocity.

since s(0) = 0
D = 0

so D dissapears.

and since s(1) = 20
(1/12) - (7/6) + (3/2) + C = 20
C = (240 + 14 - 18 - 1)/12
C = 235/12

so in short, you forgot to put a 't' with the 'C' to make it 'Ct' and you got switched values for C and D.