let b>a
b-a=14, so we can say b=14+a
ab=1800, and using b from above we get:
a(14+a)=1800
14a+a^2=1800
a^2+14a-1800=0
a^2-36a+50a-1800=0
a(a-36)+50(a-36)=0
(a+50)(a-36)=0
So there are two solutions...
a=-50 and 36 and since b=14+a the corresponding b values are:
b=-36 and 50
So the number pairs are:
-50 and -36 and
36 and 50