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Solve by completing the square: \[3x^2 - 6x - 5 = 0\]
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OK
\[3x^2-6x=5\] \[x^2-2x=5/3\] \[(x-1)^2=5/3 +1\] \[x-1=\pm \sqrt{(8/3)}\] \[x=1\pm \sqrt{8/3}\]
you failed to complete the square...
why?
completing the square means that you take b/2 and square it to get c...
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I am pretty sure that I did what the question asked
completing the square is not shown in your work. you moved the 5 to the other side and divided by three. this is not completing the square
completing the square would start like: \[{-6 \over 2}, -3^2 = 9\]
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