How do I solve by completing the square for: 1/2 n^2 + n = -1/8?

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How do I solve by completing the square for: 1/2 n^2 + n = -1/8?

Mathematics
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I got n = 1 +/- √15/4 but the answer is supposed to be: -0.13, -3.87
ohhh, never mind I think I got it now!
Step 1. multiply the equation by 2 so that the coefficient of n^2 is 1, getting: n^2 + 2n = -2/8. O.K.

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I'll stick around if you want to do the problem, I see you have figured it out.
okay so I'm getting confused: n + 1 = +/- √3/2 and n = -1 +/- √3/2
I will work it out and you can compare answers. Step 2 take half of the n coefficient (2) and square it and add it to the left side and right side. \[n ^{2}+2n + 1 = -2/8 +1\]\[(n + 1)^{2}=6/8\] We now have a perfect square on the left. Taking the square root of both sides give: n+1 =(6/8)^1/2 \[n+1 =\sqrt{6/8}\]\[n=-1\pm \sqrt{6}/\sqrt{8}\] note that \[\sqrt{8}=2\sqrt{2}\]and\[\sqrt{6}=\sqrt{3}\sqrt{2}\] The fraction now becomes\[\sqrt{3}\over 2\] so the answer is:\[n =-1\pm \sqrt{3}/2\]
Yes, the fraction becomes simpler as it is reduced to sqrt3/2
Looks like we got the same answer. Good luck with these.
but.. isn't it wrong?
why do you think it is wrong?
Do you the answer as decimal numbers?
\[-1\pm.866025403\]

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