Problem Solving

From OG 12th Ed pg. 22 #16

sqrt(3-2x) = sqrt(2x) +1
then 4x^2 =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Answer is E

My real equation is whether or not there is some set of rules that limit this question from being solved other than the OG way. For example, I solved it by:

sqrt(3-2x)^2 = (sqrt(2x)+1)^2
3-2x= 2x + 2sqrt(2x) +1
2-4x= 2 sqrt(2x)
1-2x= sqrt(2x)

This is where I decided to deviate from OG's way:

1- sqrt(2x) =2x
(1-sqrt(2x))^2= (2x)^2
1- 2sqrt(2x) + 2x= 4x^2

However, what I found out is that 1- 2sqrt(2x) + 2x does not equal to 6x-1! For example by using quadratic formula, one of the solutions for x is about 1.309.

4(1.309^2) = 6(1.309)-1

But, 1 - 2sqrt(2(1.309)) + 2(1.309) does not equal to 4(1.309)^2.

Why is that? I want to say I made a mathematical error, but I checked my work numerous times and still got the same result. What am I missing here?
Thanks!