How do you solve $x^{2} + 2 x - 3 = 0$ $x^2+2x-3=0$ by completing the square?

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Answer 1 · 2015-05-05T11:22:22

$x^{2} + 2 x - 3 = 0$ $x^2+2x-3=0$

If $x^{2} + 2 x + a^{2}$ $x^2+2x+a^2$ is a square then $a^{2} = 1$ $a^2=1$

$x^{2} + 2 x + 1 - 1 - 3 = 0$ $x^2+2x+1 -1 -3 = 0$

${(x + 1)}^{2} = 4$ $(x+1)^2 = 4$

$x + 1 = \pm \sqrt{4}$ $x+1 = +-sqrt(4)$

$x = - 1 \pm 2$ $x = -1 +-2$

$x = - 3 or x = 1$ $x=-3 " or " x=1$

How do you solve x2+2x−3=0x^2+2x-3=0 by completing the square?