Help with physics motion question (rocket)

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.70 m/s^2 . At 10.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.

How long after it was launched will the rocket fall back to the launch pad?

I found that the maximum height is 172m, but I don't know where to go from there...

The upward velocity of the rocket after 10 s = 2.70 * 10 = 27 m/s
Distance traveled in 10 s = (1/2) * 2.7 * 10^2 = 135 m

Let t = time for the rocket to reach the fround from this height of 135 m
Using s = ut + (1/2) gt^2
with s = - 135, u = 27, g = - 9.81
=> - 135 = 27t - 4.95 t^2
=> 4.95 t^2 - 27t - 135 = 0
=> t = (1/9.81) [27 + √((27)^2 + 2673)]
=> t = 8.7 s
=> time to fall back after launching
= 10 + 8.7 s
= 18.7 s.

motion to failure: assume acceleration is constant (A = 2.70 m/s/s)

v(t) = A t + 0
x(t) = (1/2) A t^2 + 0

at t=T=10s, motor fails

v(T) = A T
X(T) = (1/2) A T^2

these become the initial conditions for the second part of the problem

in the second part of the problem

a(t) = -g

v(t) = -g t + v(T) = -g t + A T
x(t) = -(1/2) g t^2 + AT t + X(T)

maximum height at time t=Q, when v(t=Q) = 0
v(t=Q) = 0 = -g Q + AT ; solve for Q

time rocket his the ground t=10+R. but because of the way your broke up time, this corresponds to
x(t=R) = 0 = -(1/2) g R^2 + AT R + X(T)
solve for R