How do you solve $x^{2} + 14 x + 20 = 0$ $x^2 + 14x + 20 = 0$ by completing the square?

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Answer 1 · 2016-08-30T19:06:02

$x = - 7 \pm \sqrt{29}$ $x = -7 +- sqrt(29)$

$1 (x^{2} + 14 x + n - n) = - 20$ $1(x^2 + 14x + n - n) = -20$

$n = {(\frac{b}{2})}^{2} = {(\frac{14}{2})}^{2} = 49$ $n = (b/2)^2 = (14/2)^2 = 49$

$1 (x^{2} + 14 x + 49 - 49) = - 20$ $1(x^2 + 14x + 49 - 49) = -20$

$1 (x^{2} + 14 x + 49) - 49 = - 20$ $1(x^2 + 14x + 49) - 49 = -20$

$1 {(x + 7)}^{2} = 29$ $1(x + 7)^2 = 29$

$(x + 7) = \pm \sqrt{29}$ $(x + 7) = +-sqrt(29)$

$x = \pm \sqrt{29} - 7$ $x = +-sqrt(29) - 7$

Hopefully this helps!

How do you solve x2+14x+20=0x^2 + 14x + 20 = 0 by completing the square?