Please help, I'm stuck at 4 AM doing math!

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Please help, I'm stuck at 4 AM doing math!

Mathematics
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Solve in the complex number system: \[x^2+3=x\]
\(x^2-x+3=0\implies (x-\dfrac{1}{2})^2=-3+(\dfrac{1}{2})^2\\(x-\dfrac{1}{2})^2=\dfrac{-11}{4}\)
Can you complete here?

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Is completing the square the only method or was I allowed to use the quadratic formula?
I can finish off from here, thanks for the help :)
sure, you can use that. This was just faster...
The quadratic equation is derived using the "completing the square" method :)
Oh really? Never really understood how the quadratic equation worked even though i'm in pre-calculus.
\[ax^2+bx+c=0\\a(x^2+\frac{b}{a}x)=-c\\a(x+\dfrac{b}{2a})^2=-c+a\dfrac{b^2}{4a^2}\\(x+\dfrac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\\(x+\dfrac{b}{2a})^2=\dfrac{b^2-4ac}{4a^2}\\(x+\dfrac{b}{2a})=\pm \sqrt{\dfrac{b^2-4ac}{4a^2}}\\(x+\dfrac{b}{2a})=\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Holy guacamole! :o *jaw drops*
Here's another explanation of deriving the quadratic formula: http://www.1728.org/quadr2.htm
pretty much the same way.
steve - did you check that answer?
the (-11/4) answer?
Yup, I got the answer right.
The answer is (-11/4)
that wasn't the answer... @lauren.ascough
I don't get that answer a=1 b=-1 c=3 right?
Yeah
x = 1 +- sq root(1 -4*1*3) / 2 x = [1 +- sq root(-11)] / 2
yeah, that is what I got.
Okay, it's just that I didn't see that answer printed anywhere.
Thanks for the answer though.
Im stuck at 6AM playing and modding skyrim Why? BECAUSE IM IN VIRTUAL SCHOOL

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