How would you complete the square with -x^2+2x-6=0?

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How would you complete the square with -x^2+2x-6=0?

Mathematics
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-x^2+2x-6 = 0 x^2-2x = -6 x^2-2x+1 = -6 +1 (x-1)^2 = -5 x - 1 = sqrt(-5) x = sqrt(-5) - 1 x = + or - i*sqrt(5) - 1
wouldn't you bring it over to the other side as a positive 6 since you're adding it?
Before you complete the square, it has to be in the form: x^2 + bx = c

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Other answers:

I just noticed you changed the signs to make the x positive, that makes more sense, thank you! does the x always have to be positive before completing the square?
Actually, let me post it a little differently so that the steps are expressed as close to the proper method as possible.
Alright, because I'll need it for future reference, thank you so much
-x^2+2x-6 = 0 -(x^2-2x +6) = 0 x^2-2x +6 = 0 x^2-2x = -6 x^2-2x+1 = -6 +1 (x-1)^2 = -5 x - 1 = sqrt(-5) x = sqrt(-5) - 1 x = + or - i*sqrt(5) - 1
There we go
for the ending, why don't you put 1 + or - i*sqrt(5)?
or isn't that generally the same concept?
x = + or - sqrt(-5)i - 1
What seems to be the trouble. You're familiar with complex numbers right?
Oh, I see
x = + or - sqrt(-5)i + 1
I'm not perfect
But could the +1 go in the front? Nobodies perfect, I'm in precalc and even my teacher makes simple mistakes. Thank you a lot though this really helped!
x = 1 + or - i*sqrt(5)?
You can write it that way too
alright, I tried working it out on my own after I saw you extracted the - from the original equation, and came to that same conclusion. thank you! really, that made complete sense!
Nice

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