Question

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Answer 1

set
\[x^2-8x-12=0\] and solve for \(x\)
since \(-8\) is even it is easiest to complete the square

Answer 2

boy was that wrong
\[x^2-8x-12=0\]
\[x^2-8x=12\]
\[(x-4)^2=12+16=28=2\sqrt{7}\]
\[x-4=\pm2\sqrt{7}\] \[x=4\pm2\sqrt{7}\]