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find the real solutions.
3+ the square root of (2x+2)=x

3+√(2x+2) = x
√(2x+2) = x-3
2x+2 = (x-3)^2
2x+2 = x^2-6x+9
x^2-8x+7 = 0
(x-1)(x-7) = 0
x = 1 or 7

check for extraneous roots:
3+√(2+2) = 3+2 = 5 nope
3+√16 = 3+4 = 7 ok

so, x=7 is the only solution

x=1 is the solution using the -√
3-√4 = 1

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