Solution: Because A is a positive integer and the original equation is a quadratic equation about X, in order to make the equation have a real root, its discriminant must be non-negative, that is, △≥0,

And △ = [2 (2a- 1)] 2-4a * 4 (a-3).

=4(4a^2-4a+ 1)- 16a^2+48a

=32a+4

Obviously, the discriminant is greater than 0.

The original equation is arranged as follows:

A (x 2+4x+4) = 12+2x, when x=-2, the two sides of the equation are not equal, so x≦-2, that is, x 2+4x+4 ≠ 0, so there is

a=( 12+2x)/(x^2+4x+4) - ①

Because x 2+4x+4 = (x+2) 2 > 0 and a is a positive integer,12+2x >; 0, and

12+2x≥x^2+4x+4

The solution is:-4 ≤ x ≤ 2; Where x≦-2.

The possible values of x are: -4, -3,-1, 0, 1, 2;

Substituting in Formula ①, the corresponding values are: 1, 6, 10, 3, 14/9,1;

A take 1, 3, 6, 10, and leave the rest.