To have both roots greater than 2, it is implied that the roots are real and not complex. Then necessarily the discriminant must be non-negative.
For this quadratic b^2-48a>=0
With a>=1 then the smallest b is 7. So b is limited to 7,8,9,10.
If b=7 then a<=1, so (a,b)=(7,1) is possible.
If b=8 then a<=1, so (a,b)=(8,1) is possible.
If b=9 then a<=1, so (a,b)=(9,1) is possible.
If b=10 then a<=2, so (a,b)=(10,1) and (10,2) are possible.
With only five cases this can be done by hand, no computer program needed.
(a,b)=(7,1) -> x^2-7x+12 -> x=3 or x=4; both greater than 2.
(a,b)=(8,1) -> x^2-8x+12 -> x=2 or x=6; one root is greater than 2 but the other root is 2, which is equal but not greater.
(a,b)=(9,1) -> x^2-9x+12 -> x=(9+/-sqrt[33])/2; one root is greater than 2 but the other root is less than 2.
(a,b)=(10,1) -> x^2-10x+12 -> x=5+/-sqrt[13]; one root is greater than 2 but the other root is less than 2.
(a,b)=(10,2) -> 2x^2-10x+12 -> x=2 or x=3; one root is greater than 2 but the other root is 2, which is equal but not greater.
So the solution set is a single pair (a,b)=(7,1). If the condition was greater than or equal to 2 then (a,b)=(8,1) and (10,2) would also be solutions.
Solution
