How do you solve the equation $3x^2-5x-8=2x-3$ by completing the square?

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Answer 1 · 2017-07-24T11:25:08

$x =1/6(7+-sqrt109)$

$3x^2 - 5x -8 =2x-3 or 3x^2 - 5x -8 -2x+3 =0$ or

$3x^2 - 7x -5 =0 or 3 (x^2-7/3x) -5$ or

$3(x^2-7/3x+(7/6)^2) - 3*49/36 -5 =0$

$3 (x-7/6)^2 - 49/12-5 or 3 (x-7/6)^2 = 109/12$ or

$(x-7/6)^2 = 109/(3*12) or (x-7/6) = +- sqrt (109/36)$ or

$x = 7/6+- sqrt (109)/6 =1/6(7+-sqrt109)$ [Ans]

How do you solve the equation 3x^2-5x-8=2x-33x^2-5x-8=2x-3 by completing the square?