2*(x+4)-((1-(3*x))^2)

First of all what was the need to use do many brackets! your question is looking more like a shop of brackets or in other words like Einsteins hair! Anyways I'll solve it

2*(x+4)-((1-(3*x))^2)

now multiply 2 by x and 4 , multiply 3 by x and just cut all those useless brackets!

= 2x + 8 - (1- 3x)^2

now here we will use an identity for the expression in brackets 

Identity is (a-b)^2= a^2 - 2ab + b^2 

now we will use this identity to solve the bracketed expression!

= 2x + 8 - 1 - 2x1x3x + (3x)^2

= 2x + 8 - 6x + 9x^2

now we wil arrange them with their alikes

= 2x - 6x + 8 + 9x^2

= -4x + 8 + 9x^2  ANSWER

2*(x+4)-((1-(3*x))^2)

$$\\(1-(3*x))^2\\
=(1-3x)^2\\
=1-6x+9x^2\\$$

2(x+4)-(1-6x+9x²)

=2x+8-1+6x-9x²

= -9x²+8x+7

We  got different answers Rosala.

Can you work out who got what wrong?

The correct answer is: -9x²+8x+7   (checked by wolfram alpha)

I think the wrong answer originated from here:

 "... 2x + 8 - (1- 3x)^2

now here we will use an identity for the expression in brackets 

Identity is (a-b)^2= a^2 - 2ab + b^2 

now we will use this identity to solve the bracketed expression!

= 2x + 8 - 1 - 2x1x3x + (3x)^2 ..."

I'll solve it like this:

2x + 8 - (1 - 3x)² =

2x + 8 - (1 - 6x + 9x²) =

2x + 8 - 1 + 6x - 9x² 

So it's mistakes made based on the minus before the paranthesis. 

Thanks Tetration. :)