Answer:
There are two sets of numbers that meet the criteria:
1) 8 and 16, or
2) -8 and -16
Explanation:
Let's call the two numbers A and B.
We are told:
1) A-B=8, and
2) A^2 + B^2 = 320
Rearrange 1) to solve for A:
A=8+B
Use this value of A in the second equation:
A^2 + B^2 = 320
(8+B)^2 + B^2 = 320
(64+ 16B + B^2) + B^2 = 320
2B^2 + 16B + 64 = 320
B^2 + 8B + 32 = 160
B^2 + 8B + 32 = 160
B^2 + 8B - 128 = 0
(B+16)(B-8)=0
B is either -16 or +8
Assuming B = 8
Then from A-B=8 we find that A = B+8 and A = 16
Does A^2 + B^2 = 320?
(16)^2 + (8)^2 = 320?
256 + 64 = 320? YES
--------------------------------
Assuming B=-16
If B=-16, then from A-B=8 we find that A = 8+(-16) and A = -8
Does A-B=8?
-8 - (-16) = 8? YES
Does A^2 + B^2 = 320?
(-8)^2 + (-16)^2 = 320?
64 + 256 = 320? YES