May 6, 2024

This can be generalized:

(Rx^2+Sx+T)^(Ax^2+Bx+C)=1

Case 1: Rx^2+Sx+T-1=0 and apply the quadratic formula.

Case 2: Rx^2+Sx+T+1=0 AND Ax^2+Bx+C-2n=0, where n is any integer, even negative.

Case 3: Ax^2+Bx+C=0 (note this is Case 2 for n=0)

To "keep things real", R^2-4S(T-1), R^2-4S(t+1), and A^2-4BC have to be greater than or equal to zero.

Note this Case 2 will always yield an infinite number of solutions.