If x # y = 3x² – y and x @ y = x + 2y, which of the following is equal to 2x # (2x @ y)?

A. 2x² + 8xy + 4y²
B. -x² + 8x – 4y
C. 10x² – 8 – 4y
D. 8x² – 4y²
E. 12x² – 2x – 2y

(E) Recall from the rules of order of operations (PEMDAS), that we always work with what is in parenthesis first. Essentially, we want to work from the inside out with this problem.

2x # (2x @ y) = 2x # (2x + 2y) = 3(2x)² – (2x + 2y) = 12x² – 2x – 2y.

The correct answer is E.
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How do you derive "12x²..." in answer choice E? Shouldn't it be "6x²..."?

Let's write the calculations in details:
1. 2x # (2x @ y) =

We apply the definition of the operation "@" to 2x and y.
2x @ y = (2x) + 2y = 2x + 2y

2. 2x # (2x @ y) = 2x # (2x + 2y) =

We use the definition of the operation "#" for 2x and (2x + 2y).
2x # (2x + 2y) = 3(2x)² – (2x + 2y) = 3 × 2² × x² – 2x – 2y
3 × 2² = 3 × 4 = 12

Therefore
2x # (2x @ y) = 2x # (2x + 2y) = 12x² – 2x – 2y