How do you solve the equation $2 x^{2} + 4 x - 5 = 7$ $2x^2+4x-5=7$ by completing the square?

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Answer 1 · 2017-08-27T10:07:40

Solution : $x = - 1 \pm \sqrt{7}$ $x = -1+- sqrt7$

$2 x^{2} + 4 x - 5 = 7 or 2 (x^{2} + 2 x) = 12$ $2x^2 +4x-5=7 or 2(x^2 +2x) =12$ or

$2 (x^{2} + 2 x + 1) = 12 + 2 or 2 {(x + 1)}^{2} = 14$ $2(x^2 +2x +1) =12+2 or 2(x + 1)^2 =14$ or

${(x + 1)}^{2} = 7 or (x + 1) = \pm \sqrt{7}$ $(x + 1)^2 =7 or (x+1) = +-sqrt7$ or

$:. x = -1+- sqrt7$

Solution : $x = -1+- sqrt7$ [Ans]

How do you solve the equation 2x^2+4x-5=72x2+4x−5=72x^2+4x-5=7 by completing the square?