The root of the equation 4^x - 3.2^(x+2) + 32 = 0

Split the second term and
rewrite 4^(x) as 2^(2x):
2^(2x) - 3 * [2^(x) * 2^2] + 32
= 0
2^(2x) - 12 * 2^(x) + 32 = 0
Let's give 2^x a value of y:
y^2 - 12y + 32 = 0
Factor it out:
(y - 8)(y - 4) = 0
y = 4, 8
Now substitute 2^x back for y
and solve for each:
2^x = 4
x = 2
2^x = 8
x = 3
----------
Overall:
x = 2, 3
option (2)

we can write the equation like this
(2^x)^2 -3 .2^x.2^2 + 32 = 0
and assume 2^x=a
equation becomes a^2-12.a+32=0
a^2- 4.a -8.a+32=0
a(a-4) - 8 (a -4) = 0
(a-4) (a-8)=0
now a - 4 = 0 and a - 8 = 0
here we get two solutions a = 4 , 8
now equate 2^x = 4 and 2^x = 8
2^x =2^2 and 2^x =2^3
hence equating both equation we get x= 2 , 3
option 2 is correct

2^2x-3*2^(x+2)+32=0

2^2x-3.2^x 2^2+32=0

2^2x-12^2x+32=0

2^x=y

y^2-12y+32=0

y^2-8y-4y+32=0

y(y-8)-4(y-8)=0

(y-8)(y-4)=0

y=8 , y=4
2^x=y
x=3 , x=2

4^x - 3.2^(x+2) + 32 = 0
2^2x - 3.(2^x * 2^2) + 32 = 0 ------(i)
Let 2^x = y
Now (i) becomes,
y^2 - 12y + 32 = 0
(y-8)(y-4) = 0
y=8 and y=4
So,2^x = 8 and 2^x = 4
x=3 and x=2
Value of x = 2 and 3 -----------option (2)

by putting the value of x we get the answer
here x=2 and 3
option 2

by rewriting the equation
2^(2x)-3[(2^x)*(2^2)]+32=0
put 2^x=y
y^2-12y+32=0
y=4,8
by substituting y=2^x
x=2 and 3

the equ. 4^x - 3.2^(x+2) + 32 can be written as
( 2^x)^2 - 3(2^x)2^2 + 32 = 0 by using law of exponent

let r=2^x then the above equa. becomes
r^2 - 3(r)2^2 +32 = 0
r^2 -12r + 32 =0
r^2 - 8r - 4r + 32 = 0
(r^2 - 8r) - ( 4r - 32) = 0
r(r - 8) - 4( r - 8 ) = 0
( r - 4 ) ( r - 8 ) = 0
=> r = 4 or r = 8

but r= 2^x
so, for r = 4 = 2^2 we've 2^2 = 2^x => x = 2

simlly, for r = 8 = 2^3 we've 2^3 = 2^x => x = 3

Hence x = 2 or x = 3
(OPTION No. 2) is the correct answer

checking
by putting x=2 into equ. 4^x - 3.2^(x+2) + 32 = 0
we've 16 - 48 + 32 = 0
=> 0 = 0
smlly, for x=3 we've 64 -96 + 32 = 0
=> 0 = 0

t^2-12t+32=0
where t=2^x
t=4 or 8
x=2 or 3

2^x=y
eq will y^2-12y+32=0
i.e. y=4 and y=8;
hence x=2 and x=3.

4^x-3.2^(x+2)+32=0
(2^x)^2-3(2^x)2^2 +32=0
a^2 -12 a +32=0
a=2^x
(a-4)(a-8)=0
a=4,8
therefore,
2^x=2^2 or 2^3
x= 2 or 3