No two intergers can solve this problem. If you're not in Algebra II yet, the answer is probably "none."
Just look at the factor pairs of 25:
1 × 25
-1 × 25
5 × 5
-5 × -5
Clearly none of those can add up to 6.
If you want the more complex answer, I'll show you how here. If you don't understand why it doesn't work, that's okay. I just want you to see that there's not an actual answer to the problem.
a+b = 6
a = 6-b
ab = 25
(6-b)b = 25
6b -b² = 25
-b² + 6b = 25
b² -6b = -25
Factor by splitting the middle.
Half of -6 is -3, (-3)² = 9. Add this to each side.
b² -6b + 9 = -16
Factor the perfect square trinomial.
(b-3)² = -16
Take the square root of each side.
b-3 = 4i
b = 3+4i
a+b = 6
a+3+4i = 6
a= 3-4i
(The "i" stands for an imaginary number, specifically, the square root of -1.)
