A ball is launched into the sky at 18 feet per second from a 18 meter tall building. The equation for the ball's height, h, at time t seconds is h = - 6t2 + 18t + 18. When will the ball strike the ground?

I believe that you meant:

h (t) = - 16t^2+18t+18 (because - 6t^2 would mean the acceleration due to gravity would be - 12ft/s^2, which certainly is not the case on Earth : P)

Anyway, the ball will hit the ground when h (t) = 0 so:

-16t^2+18t+18=0

16t^2-18t-18=0

16t^2-18t=18

t^2-9t/8=9/8

t^2-9t/8+81/256=9/8+81/256

(t-9/16) ^2=369/256

t-9/16=±√ (369/256)

t=9/16±√ (369/256)

t = (9±√369) / 16, since t>0

t = (9±3√41) / 16 seconds

t≈1.763 seconds (to nearest thousandth of a second)