let x be the larger number
& y be the small number
From the question, we have
x+y=16.......(1)
y^2-x=14..........(2)
from (1),x=16-y.......(3)
sub. (3) into (2)
y^2-(16-y)=14
y^2-16+y=14
y^2+y=14+16
y=30-y/y
y=30/y-y/y
=30/y ---1
This result tels 1 point ,30 is divisible by y and y is the integer
so there are 8 numbers can be y ------1,2,3,5,6,10,15,30
Only 5 can be y from above equation
So ,y=5
Sub y=5 into (1)
5+x=16
x=11
So , the larger number is 11
&the smaller number is 5