How do you solve the quadratic equation by completing the square: $x^{2} - 12 x + 20 = 0$ $x^2-12x+20=0$ ?

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How do you solve the quadratic equation by completing the square: $x^{2} - 12 x + 20 = 0$ $x^2-12x+20=0$ ?

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Answer 1 · 2015-07-22T12:28:15

I found:
$x_{1} = 10$ $x_1=10$
$x_{2} = 2$ $x_2=2$

Explanation:

Rearrange your equation:
$x^{2} - 12 x = - 20$ $x^2-12x=-20$
add and subtract $36$ $36$ ;
$x^{2} - 12 x + 36 - 36 = - 20$ $x^2-12xcolor(red)(+36-36)=-20$
$x^{2} - 12 x + 36 = - 20 + 36$ $x^2-12x+36=-20+36$
so you can write:
${(x - 6)}^{2} = 16$ $color(blue)((x-6)^2=16)$
square root both sides:
$x - 6 = \pm \sqrt{16} = \pm 4$ $x-6=+-sqrt(16)=+-4$
so you end up with two solutions:
$x_{1} = 6 + 4 = 10$ $x_1=6+4=10$
$x_{2} = 6 - 4 = 2$ $x_2=6-4=2$

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How do you solve the quadratic equation by completing the square: $x^{2} - 12 x + 20 = 0$ $x^2-12x+20=0$ ?

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How do you solve the quadratic equation by completing the square: $x^{2} - 12 x + 20 = 0$ $x^2-12x+20=0$ ?