What multiplication problem can give you negative 18, but when subtracted, gives you a negative 1?

a*b=-18 
a-b=-1 

which we rewrite as 
a=b-1 

then we substitute 
(b-1)b=-18 
Now we solve for b 

b^2-b+18=0 

which has the following solution 
 
So as you can see there is no real pair of numbers that can do both at the same time but if we allow the numbers to be complex, then it is possible 

:)