How do you complete the square $3x^2-6x-1=0$ ?

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Answer 1 · 2015-02-04T18:22:31

The answer is:

$3(x^2-2x)-1=0rArr3(x^2-2x+1-1)-1=0rArr$

$3(x^2-2x+1)-3-1=0rArr3(x-1)^2-4=0rArr$

$(x-1)^2=4/3rArrx-1=+-sqrt(4/3)=0rArr$

$x=1+-2/sqrt3rArrx=1+-(2sqrt3)/3rArr$

$x=(3+-2sqrt3)/3$ .

How do you complete the square 3x^2-6x-1=03x^2-6x-1=0?