When you put the given numbers (v=122, c=99) into the vertical motion formula, you get
0 = -16t² + 122t + 99
Solving that using the quadratic formula for a=-16, b=122, c=99, you get
t = (-b±√(b²-4ac))/(2a)
t = (-122 ±√(122²-4·(-16)·99))/(2·(-16))
t = (122 ±√21220)/32
t = 3.8125 ± √20.72265625
t ≈ -0.7 or 8.4
The appropriate choice is ...
C. 0 = -16t² + 122t + 99; 8.4 s