Подробное решение пожулайста

$2 \times {4}^{x + 1} - 33 \times {2}^{x} + 4 = 0$
$2 \times 4 \times {2}^{2x} - 33 \times {2}^{x} + 4 = 0$
Пусть
$t = {2}^{x}$
Тогда:
8t^2-33t+4 =0
D = 1089 - 4*8*4 = 961 = 31^2
t1=(33+31)/16 = 4
t2=(33-31)/16 = 2/16 = 1/8

${2}^{x} = 4$
x1=2
${2}^{x} = \frac{1}{8}$

x2= -3

2) 9^(x+1)-28*3^x+3=0
9*3^(2x) - 28*3^x + 3 = 0
3^x = t
9t^2 -28t+3=0
D = 784 - 4*9*3 = 676 = 26^2
t1 = (28+26)/18 = 3
t2 = (28-26)/18 = 2/18 = 1/9

3^x=3
x1= 1

3^x=1/9
x2 = -2