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How do you solve $x^2-8x-1=8$ by completing the square?

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Jun 15, 2018

$x^2-8x-1=8$

$x^2-8x-9=0$

$(x-4)^2-16-9=0$

$(x-4)^2-25=0$

$(x-4)^2=25$

$(x-4)=\pm5$

$x-4=5 or x-4=-5$

$x=9 or x=-1$

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