How do you solve $0=3x^2-11x+6$ by completing the square?

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Answer 1 · 2015-06-05T19:05:00

$0=3x^2-11x+6$

$=3(x^2-11/3x+2)$

$=3(x^2-11/3x+121/36-121/36+2)$

$=3((x-11/6)^2-121/36+72/36)$

$=3((x-11/6)^2-49/36)$

Divide both sides by $3$ to get:

$(x-11/6)^2-49/36 = 0$

Add $49/36$ to both sides to get:

$(x-11/6)^2 = 49/36$

Take square root of both sides to get:

$x-11/6 = +-sqrt(49/36) = +-7/6$

Add $11/6$ to both sides to get:

$x = 11/6+-7/6 = (11+-7)/6$

That is $x = 3$ or $x = 2/3$

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