3.086 seconds.
First, convert 250 km/h into m/s by multiplying by 1000 (to convert km to m) and dividing by 3600 to convert per hour to per second. Doing 250 * 1000 / 3600 gives you a speed of 69.4444 m/s.
Now under constant acceleration, the formula for distance is:
1/2 A T^2
Since A = -8.25 (negative since it's deceleration) we get
1/2 * (-8.25) * T^2
giving
-4.125 T^2
The total distance traveled then becomes the original velocity multiplied by the time plus the distance for the acceleration, giving
69.4444 T - 4.125 T^2 = 175
Doing a bit of reordering, we get a standard looking quadratic equation
-4.125 T^2 + 69.4444 T - 175 = 0
Plugging the values of A = -4.125, B = 69.4444, C=-175 into the quadratic formula gives you two solutions.
First solution is 3.086 seconds
and the second solution is 13.750 seconds.
The physical interpretation of those 2 answers is that 3.086 seconds after the reverse thrust is applied, the vehicle is 175 meters from the spot where the thrust was applied. The vehicle continues to decelerate until it comes to a complete stop. Then the vehicle goes into reverse and after a total of 13.75 seconds, once again is at a point 175 meters from where it had applied the reverse thrust. We can use that information to double check our answer. Namely at the mid point between 3.086 seconds and 13.75 seconds, the rocket car should be motionless. Let's check if that's correct.
(3.086 + 13.75) / 2 = 8.418 seconds
Given a velocity of 69.44444 m/s and a deceleration of 8.25 m/s^2, the vehicle should stop after
69.44444 / 8.25 = 8.418 seconds
Since those two independent calculations of when the vehicle would stop the match, it's a good check that the math is correct.
