t = 0.005x² + 0.002x
9 = 0.005x² + 0.002x
0 = 0.005x² + 0.002x - 9
x = -(0.002) ± √((0.002)² - 4(0.005)(-9))
2(0.005)
x = -0.002 ± √(0.000004 + 0.18)
0.01
x = -0.002 ± √(0.180004)
0.01
x = -0.002 ± 2√(0.053911)
0.01
x = -0.2 ± 200√(0.053911)
The number of objects that is required to keep the computer busy in exactly 9 seconds is -0.2 ± 200√(0.053911).