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

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Answer 1 · 2015-04-29T19:59:37

Move the constant term to the right side of the equation, then add whatever quantity is necessary to both sides so the left side is a square

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

$x^{2} + 2 x = 7$ $x^2+2x = 7$

If the first two terms of ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2 +2ax +a^2$
are $x^{2} and 2 x$ $x^2 and 2x$
then $a = 1$ $a=1$

$x^{2} + 2 x + 1 = 7 + 1$ $x^2+2x+1 = 7+1$

${(x + 1)}^{2} = 8$ $(x+1)^2 = 8$

$x + 1 = \pm \sqrt{8}$ $x+1 = +-sqrt(8)$

$x = - 1 \pm 2 \sqrt{2}$ $x = -1+-2sqrt(2)$

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