3*9^x + 11*3^x = 4
3*3^2x + 11*3^x - 4 = 0
3^x = t ; t>0
3t^2 + 11t - 4 = 0
D = b^2 - 4*a*c
D = 11^2 + 4*3*4 = 121 + 48 = 169
t1,2 = (-b ± √D) / 2*a
t1 = (-11 + 13)/6 = 2/6 = 1/3
t2 = (-11 - 13)/6 = -24/6 = -4 - лишний корень, т.к. t>0
3^x = 1/3
x = -1