This is a multiple choice question.

Question

What is the solution of the equation 2(x+2)^2-4=28
1)2 only
2)6 only
3)-2 and 6
4)2 and -6

I did (2x+4)(2x+4)-4=28
4x^2+8x+8x+16-4=28
4x^2+16x+12=28
4x^2+16x=16
This is where I get stuck, can you help? thank you

Answer 1

On the first line, the 2 should get distributed only across the first set of brackets:
(2x+4)(x+2)-4=28
Work it down again similar to what you did, then get all your terms to one side of the equation. Solve by factoring (if possible) or by using the quadratic formula.

Answer 2

ok, corrected is
2x^2+4x+4x+8-4=28
2x^2+8x+4=28
2x^2 +8x-24=0
none of them still work

Answer 3

That's correct.
Then,
2x^2+8x-24=0
2(x^2+4x-12)=0
2(x+6)(x-2)=0
x = -6, x = 2