How do you solve the equation $2x^2-7x+12=0$ by completing the square?

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Answer 1 · 2018-05-11T12:34:20

Solution: $x = 1.75 +- 1/4 sqrt(47) i$

$2 x^2 -7 x +12 =0 or 2 x^2 -7 x = -12$ or

$2 (x^2 -7/2 x )= -12$ or

$2 (x^2 -7/2 x +49/16 )= 49/8-12$ or

$2 (x -7/4)^2 = -47/8$ or

$(x -7/4)^2 = -47/16$ or

$(x -7/4)= +- sqrt(-47/16)$ or

$(x -7/4)= +- sqrt(47i^2)/4 [i^2=-1]$

$x = 7/4+- 1/4sqrt(47) i$ or

$x = 1.75 +- 1/4sqrt(47) i$

Solution: $x = 1.75 +- 1/4 sqrt(47) i$ [Ans]

How do you solve the equation 2x^2-7x+12=02x^2-7x+12=0 by completing the square?