Edwin's solution:

You'll eventually get the answer that way, but it'll take
you a long long time.  Surely your teacher expects you to
have a graphing calculator, such as a TI-83 or 84.  You 
must get one and learn to use it, for this is a graphing 
calculator problem.

If you have a TI-83 or TI-84

You have to use X instead of t

Press the WINDOW key

Type these values instead of
what you find there:

WINDOW
Xmin=-1
Xmax=25
Xscl=1
Ymin=-100
Ymax=1000
Yscl=100
Xres=1

Press Y= and enter -.5X^3+11.7X^2-21.5X+218.6 next to \Y1=

so the top line reads \Y1=-.5X^3+11.7X^2-21.5X+218.6 next to \Y1=
(it will scroll onto the next line since there's not enough room
for it all to fit on one line) 

Press GRAPH

You'll see a graph.

Press TRACE

Then use your left and right arrows to trace the
curve all the way to the right end.  As you move the
cursor along the curve, the x and y coordinates
of the point it's on are given at the bottom of the
screen, to many decimal places.   

Now when you've mover the cursor to the bottom right of 
the curve you see that it goes negative when you get to 
about 22.4. And it is not valid when it goes negative. 

When you substitute in 22 you get 84.5, and when you 
substitute in 23 you get -170.1.

You can even find the value where the curve becomes
zero if you know how to use a graphing calculator.   

it is t = 22.351309 years after 1989

So it is no longer valid after 22.351309 years after 1989.
By the way, that works out to be May 18, 2011 at 6:24 PM.  
That is the instant when there will be 0 quadruplet births 
according to the modeling formula.

Edwin