Differential Equations Help!

Im not sure if i am solving this question right and if someone could point me in the right direction that would be awesome!

dP/dt=0.3(1-(P/200))((P/50)-1)P

a)For what values of P is the population in equilibrium?
-Is equilibrium the solution to the D.E?
b) For what values of P is the population increasing?
-Is this when the D.E. is concave up?
c) For what values of P is the population decreasing?
-is this when the D.E. is concave down?
Thanks sooooo much!

a) Find when dP/dt = 0

dP/dt = 0
0 = 0.3 * (1 - (P/200)) * ((P/50) - 1) * P

0 = (1 - (P/200))
P/200 = 1
P = 200

0 = (P/50) - 1
1 = P/50
P = 50

P = 0

When the population is at 0 , 50 , and 200

b)
Find when dP/dt > 0

dP/dt = 0.3 * (1 - (P/200)) * ((P/50) - 1) * P
dP/dt = 0.3 * ((200 - P)/200) * ((P - 50)/50) * P
dP/dt = (3/10) * (1/200) * (1/50) * (200 - P) * (P - 50) * P
dP/dt = (3/10) * (1/10000) * (200 - P) * (-1) * (50P - P^2)
dP/dt = (-3/100000) * (10000P - 200P^2 - 50P^2 + P^3)
dP/dt = (-3/100000) * (P^3 - 250P^2 + 10000P)

This is a cubic equation that will have a negative leading coefficient. This means that its end behaviors will have it move from Q2 to Q4. We know that dP/dt = 0 when P = 0 , 50 and 200. This means that we'll have 4 domains

-inf < P < 0
0 < P < 50
50 < P < 200
200 < P < infinity

dP/dt is positive from -inf < P < 0 and 50 < P < 200 (for further reference, this means that dP/dt < 0 from 0 < P < 50 and 200 < P < infinity, remember that for portion c).

Now, we need to find the derivative

dP/dt = (-3/100000) * (P^3 - 250P^2 + 10000P)
d2P/dt^2 = (-3/100000) * (3P^2 - 500P + 10000)

Find when d2P / dt^2 = 0

0 = 3P^2 - 500P + 10000
P = (500 +/- sqrt(250000 - 12 * 10000)) / 6
P = (500 +/- sqrt(250000 - 120000)) / 6