9^x+2 - 6* 3^x+1= 0 find x

Answer 1

If you let u=3^x then the equation becomes (using parentheses to clarify matters)

9^(x+2) - 6*3^(x+1) = 0
9^2*9^x - 6*3*3^x = 0
81u^2-18u+1=0
(9u-1)^2 = 0
u = 1/9
So, 3^x = 1/9 = 3^-2
x = -2

Answer 2

Actually how do you get +1 after81u^2-18u +1 I can't understand