For those of you who do not have a graphing calculator, or where graphing calculators are not allowed in exams, here's how we can solve the problem.
Let the smaller of the positive numbers be p, and the other must be (p+7) since the difference is 7.
The sum of squares equals 289 can be expressed as
p^2+(p+7)^2=289
expand
p^2+7^2+14p+p^2=289
2p^2+14p-240=0
p^2+7p-120=0
Factor and solve
(p-8)(p+15)=0
which means p-8=0 or p+15=0 by the zero product theorem giving solutions of p=8 or p=-15 (rejected because p must be a positive number)
So we conclude that the two numbers are 8 and 8+7=15.
