Help differentiation problem

The distance, S cm of a particle P from its fixed point O after T seconds, is given by S=T^2-8T+12

A) Find the speed an motion of when T = 3
B) Find the time when the particle is instantaneous at rest
C) Find the speed an direction of motion when the particle passes through O

A) dS/dT is velocity, so dS/dT =2T-8,
so dS/dT at T=3, -2
b) dS/dT=0 when at rest so 2T-8=0, T=4 secs
c) S=0, when passing through O
T^2-8T+12=0
(T-2)(T-6)=, so T =2 and 6 secs when it passes through O.

A)
Speed = Distance First Derivative
Speed(3) = Distance'(3) = 2T-8 |T=3 = 2*3-8 = 6-8 = -2 cm/s
(Negative sign means it goes away from the point O.)

B)
Particle is instantaneously at rest means its speed is zero at that time, solve the following:
Speed(T) = 0 --> 2T-8 = 0 --> 2T=8 --> T=4 s

C)
The particle will pass through the point O when the distance from O is zero, this means the following:
Distance(T) = 0 --> T^2-8T+12 = 0 --> (T-2)(T-6) = 0 --> Time will be either 2s or 6s.
Speed(2) = 2T-8 | T=2 = 2*2-8 = 4-8 = -4s it means the speed will be 4s and the direction is away from O.
Speed(6) = 2T-8 | T=6 = 2*6-8 = 12-8 = 4s it means the speed will be 4s and the direction is to the point O.

A) diferenciate S
2T-8=V(speed)
replace T wid 3
(2*3)-8= -2m/s

B) equate S=t^2-8t+12 to zero nd solve it,

C) sorry..no idea!

hope i helped!