How do you solve $x^2 - 6x =13$ by completing the square?

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Answer 1 · 2018-03-24T05:36:03

$x=3+sqrt22$
$x=3-sqrt22$

Given -

$x^2-6x=13$

$x^2-6x+9=13+9$

$(x-3)^2=22$

$x-3=+-sqrt22$

$x=+-sqrt22 +3$

$x=3+sqrt22$
$x=3-sqrt22$

Answer 2 · 2018-03-24T05:44:30

$x= sqrt22+3 or -sqrt22+3$

$x^2 -6x =13$

$x^2 -6x -13=0$

Multiply the equation by 4

$4(x^2 -6x -13=0)$

$4x^2 -24x-52=0$

$(2x)^2 -24x-52=0$

$(2x)^2-2×(2x)×(6)+6^2-6^2-52=0$

$(2x-6)^2 -36-52 =0$

$(2x-6)^2 -88=0$

$(2x-6)^2=88$

$2x-6= +-sqrt 88$

$2x-6=sqrt 88 or 2x-6=-sqrt88$

$2x-6=2sqrt22 or 2x-6=-2sqrt22$

$x= sqrt22+3 or x= -sqrt22+3$

How do you solve x^2 - 6x =13 x^2 - 6x =13 by completing the square?