How do you solve $x^{2} - 14 x + 45 = 0$ $x^2-14x+45=0$ by completing the square?

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Answer 1 · 2015-05-03T23:00:52

$x^{2} - 14 x + 45 = 0$ $x^2-14x+45 =0$

$(x^{2} - 14 x + 7^{2}) - 7^{2} + 45 = 0$ $(x^2-14x+7^2) -7^2 +45 = 0$

${(x - 7)}^{2} = 4$ $(x-7)^2 = 4$

$x - 7 = \pm 2$ $x-7 = +-2$

$x = 7 \pm 2$ $x= 7+-2$

$x = 9$ $x=9$ or $x = 5$ $x=5$

How do you solve x2−14x+45=0x^2-14x+45=0 by completing the square?