SOLUTION: I need to solve x2(this is x squared)+2x-4=0 I tried to solve by adding 4 to both sides to get x squared +2x=4 Then I divided 2x by 2 and squaring to get 1 then I added one to b

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I need to solve x2(this is x squared)+2x-4=0 I tried to solve by adding 4 to both sides to get x squared +2x=4 Then I divided 2x by 2 and squaring to get 1 then I added one to b      Log On


   

Question 447911: I need to solve x2(this is x squared)+2x-4=0
I tried to solve by adding 4 to both sides to get x squared +2x=4
Then I divided 2x by 2 and squaring to get 1
then I added one to both sides to get X squared +2x+1=5
Then I factored to get (x+1)squared=5
Then I took the square root of both sides to get x+1=sq root of 5
Then I subtracted one from both sides to get
x=-1+sq root of 5 and x=-1-sq root of 5
Is this correct?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi, Your Answer is correct, please review below

 x^2+2x    + 1 = 5  |completing the square: x^2 + 2x = (x+1)^2 - 1

 %28x%2B1%29%5E2-1+%2B1+=+5

    (x+1)^2 = 5

     x+1 = ± sqrt%285%29

      x = -1 ± sqrt%285%29