How do you solve 6-3[2x+6]=0 whats the answer?

6 - (3*2)x - (3*6) = 0

6 - 6x - 18 = 0

-12 = 6x

x = -2

6 - 6x - 18 = 0
- 12 = 6x
x = - 2

u take the -3 and give it to the 2x and tha 6 times it by each of them u should get
6 -6x -18=0
then u hav to get x by itself. the 18 is subtracting to get 0 so add it to -18 to cancel it and add 18 to 0 so its like u never subtracted it.
6 -6x=18
then the 6 is a positive so ur going to subtract it from itself and 18
-6x=12
now the -6 is timsed against the x so divide -6 against -6x to cancel the -6 then divide -6 by 12
x= -2

6 - 3(2x + 6) = 0
6 - 3*2x - 3*6 = 0
6 - 6x - 18 = 0
-6x = -6 + 18
-6x = 12
x = 12/-6
x = -2

6-6x-18=0 -6x-12=0 -6x=12 x=12/-6=-2

6-3[2x+6]=0
3[2x+6]=0
6x+18=0
(subtract 18 from both sides)
6x=-18
(divided by 6)
x=-3

6 - 3 (2x + 6) = 0
6 - 6x - 18 = 0
-6x - 12 = 0
-6x = 12
(-6x = 12)/-6
x = -2

6 - 3 [2x + 6] = 0
subtract 6 from both sides
- 3 [2x + 6] = -6
divide both sides by - 3
[2x + 6] = 2
remove brackets and subtract 6 from both sides
2x = -4
divide both sides by 2
x = -2