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How do you solve by completing the square for $2 x^{2} - 4 x + 5 = 0$ $2x^2 - 4x + 5 = 0$ ?

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Jun 3, 2015

$2 x^{2} - 4 x + 5 = 0$ $2x^2-4x+5 = 0$

$XXXX$ $color(white)("XXXX")$ $2 (x^{2} - 2 x) = - 5$ $2(x^2-2x) = -5$

$XXXX$ $color(white)("XXXX")$ $2 (x^{2} - 2 x + 1) = - 5 + 2$ $2(x^2-2x+1) = -5 +2$

$XXXX$ $color(white)("XXXX")$ ${(x - 1)}^{2} = - \frac{3}{2}$ $(x-1)^2 = -3/2$

$XXXX$ $color(white)("XXXX")$ $x - 1 = \pm \sqrt{- \frac{3}{2}}$ $x-1 = +-sqrt(-3/2)$

note: there will be no Real solutions
$XXXX$ $color(white)("XXXX")$ $x = 1 \pm \sqrt{\frac{3}{2}} i$ $x = 1+-sqrt(3/2)i$

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