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

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Answer 1 · 2017-09-03T11:15:26

Solution: $x = (3 +- 2sqrt3)/3$

$3x^2-6x-1 =0 or 3(x^2-2x)= 1$ or

$3(x^2-2x+1)= 1 +3 or 3(x-1)^2 = 4$ or

$(x-1)^2 = 4/3 or x-1 = +-sqrt (4/3)$ or

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

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

Solution: $x = (3 +- 2sqrt3)/3$
[Ans]

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