SOLUTION: Could you please help me. What is the solution of the equation 2(x+2)^2-4=28 when solved for x.

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Question 1062214: Could you please help me. What is the solution of the equation 2(x+2)^2-4=28 when solved for x.
Found 3 solutions by math_helper, ikleyn, MathTherapy:
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!




(divided previous line by 2, both sides)

x=2 and x=-6
—
Check:
x=2: 2(x+2)^2 - 4 = 2(4)^2 - 4 = 2(16) - 4 = 32-4 = 28 (ok)
x=-6 2(x+2)^2 - 4= 2(-6+2)^2 - 4 = 2(16)- 4 = 32-4 = 28 (ok)
—
Answer:
x = 2 and
x = -6 both solve the equation
—
Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
Could you please help me. What is the solution of the equation 2(x+2)^2-4=28 when solved for x.
~~~~~~~~~~~~~~~~~~~~~~ 

 = 28,

 = 28 + 4 = 32,

 =  = 16,

x + 2 = +/- = +/-4.

x = -2 + 4 = 2  OR  x = -2 - 4 = -6.

Answer. The solutions are -6 and 2.


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Could you please help me. What is the solution of the equation 2(x+2)^2-4=28 when solved for x. 






 ----- Dividing by 2

 ------ Taking the square root of each side





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