text: solving an equation.? http://img171.imageshack.us/img171/7404/project1zl9.jpg i solved it on my calculator and i got "4" answers. x= -.414, x = 2.414, and two answers which were close to "1" x  = 1.0000000~ and x=.9999999~ i tried using u-subsitution but i only got up to a certain point. then i tried another way and i proved that (x^2-2x-1)=(x^2-2x-1). <_< we have: (2+√3)^(x^2 - 2x + 1) + (2-√3)^(x^2 - 2x - 1) = 4/(2-√3)  (2-√3) [(2+√3)^(x^2 - 2x + 1) + (2-√3)^(x^2 - 2x - 1)] = 4 (2-√3)(2+√3)^(x^2 - 2x + 1) + (2-√3)(2-√3)^(x^2 - 2x - 1) = 4 (2-√3)(2+√3)(2+√3)^(x^2 - 2x) + (2-√3)^(x^2 - 2x) = 4 and (2-√3)(2+√3) = 1, so we have:  (2+√3)^(x^2 - 2x) + (2-√3)^(x^2 - 2x) = 4 again, (2-√3)(2+√3) = 1 so (2-√3) = 1 /(2+√3)  substituting,  (2+√3)^(x^2 - 2x) + 1/(2+√3)^(x^2 - 2x) = 4 let y = (2+√3)^(x^2 - 2x) so y + 1/y = 4 or y^2 - 4y + 1 = 0 y can then be found to be 2 ± √3  so (2+√3)^(x^2 - 2x) = 2 ± √3 if (2+√3)^(x^2 - 2x) = 2 + √3, then x^2 - 2x = 1 if (2+√3)^(x^2 - 2x) = 2 - √3 = (2 + √3)^(-1), then x^2 - 2x = -1  solving these for x, we find x = 1 ± √2 or the double root x = 1