The given quadratic is in vertex form, $2(x-4)^2 + 8 = 2(x^2 -8x + 16) + 8 = 2x^2 - 16x + 40$.
To express $3ax^2 + 3bx + 3c = 3(2x^2 - 16x + 40)$ in the form $n(x-h)^2 + k$,
we factor out the constant multiple of 3: $3(2x^2 - 16x + 40) = 6(x^2 - 8x + 20)$.
Next, we complete the square: $6(x^2 - 8x + 20) = 6(x^2 - 8x + 16) + 6(20-16) = 6(x-4)^2 + 24$.
Therefore, $h=4$, and the expression in the form $n(x-h)^2 + k$ is $6(x-4)^2 + 24$.
The answer is: 4
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
If the quadratic $ax^2 + bx + c$ is in the form $2(x - 4)^2 + 8$, what is the value of $h$ when the quadratic $3ax^2 + 3bx + 3c$ is expressed in the form $n(x - h)^2 + k$?

### Answer:
