A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 108 feet per second from an initial height of 102 feet off the ground, then the height of the projectile, $\displaystyle {h}$, in feet, $\displaystyle {t}$ seconds after it's shot is given by the equation:
$\displaystyle {h}=-{16}{t}^{{2}}+{108}{t}+{102}$

Find the two points in time when the object is 141 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).

Answer: The object is 141 feet off the ground at the following times:
Enter your two answers separated by a comma.