SOLUTION: Please check to see if I am doing this the correct way.
SOLVE AND CHECK
~(4x+4)+12=0
~(4x+4)=-12
(~(4x+4))^(2)=(-12)^(2)
(~(4(x+1)))^(2)=(-12)^(2)
(2)^(2)=(-12)^(2
Algebra ->
Square-cubic-other-roots
-> SOLUTION: Please check to see if I am doing this the correct way.
SOLVE AND CHECK
~(4x+4)+12=0
~(4x+4)=-12
(~(4x+4))^(2)=(-12)^(2)
(~(4(x+1)))^(2)=(-12)^(2)
(2)^(2)=(-12)^(2
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Question 98223: Please check to see if I am doing this the correct way.
SOLVE AND CHECK
~(4x+4)+12=0
~(4x+4)=-12
(~(4x+4))^(2)=(-12)^(2)
(~(4(x+1)))^(2)=(-12)^(2)
(2)^(2)=(-12)^(2)
(2~(x+1))^(2)=(-12)^(2)
4(x+1)=(-12)^(2)
(4x+4)=(-12)^(2)
4x+4=(-12)^(2)
4x+4=(144)
4x+4=144
4x=140
x=35
No Solution
Thank you Found 2 solutions by benni1013, mathslover:Answer by benni1013(206) (Show Source):
You can put this solution on YOUR website! Hold on and back the heck up: Here is how you do this problem.
(4x+4)+12=0
(4x+4)=-12. You right up to this point now pay attention
4(x+1)=-12. See what I did I made a GCF out the quantity inside the parentheses.
x+1=-3
x=-4
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