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

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Answer 1 · 2018-05-11T12:34:20

Solution: $x = 1.75 \pm \frac{1}{4} \sqrt{47} i$ $x = 1.75 +- 1/4 sqrt(47) i$

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

$2 (x^{2} - \frac{7}{2} x) = - 12$ $2 (x^2 -7/2 x )= -12$ or

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

$2 {(x - \frac{7}{4})}^{2} = - \frac{47}{8}$ $2 (x -7/4)^2 = -47/8$ or

${(x - \frac{7}{4})}^{2} = - \frac{47}{16}$ $(x -7/4)^2 = -47/16$ or

$(x - \frac{7}{4}) = \pm \sqrt{- \frac{47}{16}}$ $(x -7/4)= +- sqrt(-47/16)$ or

$(x - \frac{7}{4}) = \pm \frac{\sqrt{47 i^{2}}}{4} [i^{2} = - 1]$ $(x -7/4)= +- sqrt(47i^2)/4 [i^2=-1]$

$x = \frac{7}{4} \pm \frac{1}{4} \sqrt{47} i$ $x = 7/4+- 1/4sqrt(47) i$ or

$x = 1.75 \pm \frac{1}{4} \sqrt{47} i$ $x = 1.75 +- 1/4sqrt(47) i$

Solution: $x = 1.75 \pm \frac{1}{4} \sqrt{47} i$ $x = 1.75 +- 1/4 sqrt(47) i$ [Ans]

How do you solve the equation 2x2−7x+12=02x^2-7x+12=0 by completing the square?