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
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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)
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Answer:
x = 2 and
x = -6 both solve the equation
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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.
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= 28,
= 28 + 4 = 32,
= = 16,
x + 2 = +/- = +/-4.
x = -2 + 4 = 2 OR x = -2 - 4 = -6.
Answer. The solutions are -6 and 2.
