Find all real solutions of (x^2 +4x +3)^(x^2 -6x+8) = 1

Rewrite as a^b = 1, where a = x^2+4x+3 and b = x^2-6x+8List the cases for which a^b = 1Case 1: a = 1:So x^2+4x+3=1 => x^2+4x+2=0 => x=-2 +/- sqrt(2)Case 2: b=0, a ≠ 0:b=(x-4)(x-2) => x = 2, x=4Check that at neither of these values satisfy a=0Case 3: a = -1, b is even:So x^2+4x+3=-1 => x^2+4x+4=0 => x=-2