How do i factorise?

You have already factorised, the function is already expressed as a product of factors which are linear.

If you are wishing to solve for x:

Divide by 2

(y+6)^2 = 64

Square root (dont forget about the negative root too, lots of people have by the looks of things)

y+6 = +- 8
y=-14 or 2

So there are two sols, y=-14 or y=2

2(y + 6)(y + 6) = 128
(y + 6)(y + 6) = 128/2
y*y + 6*y + y*6 + 6*6 = 64
y^2 + 6y + 6y + 36 - 64 = 0
y^2 + 12y - 28 = 0
y^2 + 14y - 2y - 28 = 0
(y^2 + 14y) - (2y + 28) = 0
y(y + 14) - 2(y + 14) = 0
(y + 14)(y - 2) = 0

y + 14 = 0
y = -14

y - 2 = 0
y = 2

∴ y = -14 , 2

That's not factorising. It's already been factorised.
Factorising is putting numbers etc. back into the brackets.
I think you need to expand those brackets.

What?

(y+6)^2 = 64

y+6=+8 or -8

y=2 or -14

(y + 6)² = 64
y + 6 = ± 8
y = 2 , y = - 14

2(x+6)(x+6) = 128
(x+6)^2 = 64
x+6 = ±8
x = -6±8 = {2, -14}

Which can be rewritten as (x-2)(x+14)=0 for very obvious reasons.

2(y+6)(y+6) = 128
(y+6)(y+6) = 64
y^2 + 12y + 36 = 64
y^2 +12y - 28 = 0
(y+14)(y-2) = 0

Done!!!

2(y+6)(y+6) = 128

2(y + 6)^2 = 128

(y + 6)^2 = 64

(y + 6) = 8

y = 2

Do you want to "factorize" or solve for x?
If it's the latter, then

Divide by 2.

(y+6)(y+6) = 64

Take square root
y + 6 = 8

Subtract 6
y = 2