How do you solve $x^{2} + 20 x + 104 = 0$ $x^2+20x+104=0$ by completing the square?

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Answer 1 · 2016-08-13T17:35:29

$x = - 10 \pm 2 i$ $x = -10 +- 2i$

Move the constant term to the RHS.

$x^{2} + 20 x = - 104$ $x^2 + 20x = -104$

Add the square of half the coefficient of the $x$ $x$ term to both sides:

$x^{2} + 20 x + 10^{2} = - 104 + 10^{2}$ $x^2 + 20x + color(red)(10^2) = -104 + color(red)(10^2)$

This becomes:

${(x + 10)}^{2} = - 104 + 100$ $(x+10)^2 = -104+100$

${(x + 10)}^{2} = - 4$ $(x+10)^2 = -4$

Take square roots of both sides.

$x + 10 = \pm \sqrt{- 4} = \pm \sqrt{4 i^{2}} = \pm 2 i$ $x+10 = +-sqrt(-4) = +-sqrt(4i^2) = +-2i$

$x = - 10 \pm 2 i$ $x = -10 +- 2i$

How do you solve x^2+20x+104=0x2+20x+104=0x^2+20x+104=0 by completing the square?