What time does the grinding wheel stop spinning? (constant ang. accel.)

[physx_420]

Homework Statement

At t=0, a grinding wheel has omega=24.0rad/s, alpha=30.0rad/s^2 until a circuit breaker trips at t=2s. From then on it turns through 432rad as it coasts to a stop at constant alpha.
a) through what total angle did the wheel turn between t=0 and the time it stopped?
b) what time did it stop?
c) what was its acceleration as it slowed down?

Homework Equations

theta-f.=theta-in. + omega*t + 1/2*alpha*t^2
omega-f.^2=omega-in.^2 + 2*alpha*(theta-f. - theta-in.)
[-B+sqrt(b^2 - 4AC)]/2A

The Attempt at a Solution

for a) I solved theta-f.(2) to get 540 radians.
for b) I plugged in the final value of theta and moved everything over so I could use the quadratic equation to get t=.5 sec, but the answer is 12.3 s...I'm not sure where I went wrong. Help please

 

[lippyka]

?for c) I used the equation omega-f.^2=omega-in.^2 + 2*alpha*(theta-f. - theta-in.), and plugged in the initial omega and theta, and substituted t for .5, but that gave me a negative value for alpha.