Question 268286
An astronaut on the moon throws a baseball upward. The astronaut is 6ft., 6in. tall, and the initial velocity of the ball is 30 ft. per sec. The height S of the ball in feet is given by the equation
S= -2.7t^2 + 30t + 6.5
where t is the number of seconds after the ball was thrown. 
A.) After how many seconds is the ball 12ft above the moon's surface? Round to the nearest hundreth. 
I know I have to look for the seconds, which is t so then I have to solve for S, the height. This is have I've got so far:
12 = 2.7t^2 + 30t + 6.5 *** It's -2.7
-12 -12
------------------------
0 = -2.7t^2 + 30t - 5.5
*[invoke solve_quadratic_equation -2.7,30,-5.5]
---------------
Notice there are 2 numbers, the smallest is going up, the other coming back down.
-------------
B.)How many seconds will it take for the ball to return to the surface? Round to the nearest hundredth.
It's at the surface when h = 0
-2.7t^2 + 30t + 6.5 = 0
*[invoke solve_quadratic_equation -2.7,30,6.5]
------------
Ignore the negative number.
t = 11.32 seconds

So on the second part of the equation I have to look for the time, so that means that I have to solve for the height...