How do you solve $x^{2} - 10 x = 13$ $x^2- 10x= 13$ by completing the square?

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Answer 1 · 2016-09-11T18:01:12

$a x^{2} + b x + \frac{1}{2} b^{2}$ $ax^2+bx+1/2b^2$

$x^{2} - 10 x = 13, b = 10, \frac{1}{2} b^{2} = 5^{2} = 25$ $x^2-10x=13, b=10, 1/2b^2=5^2=25$

Add 25 to both sides

$x^{2} - 10 x + 25 = 38$ $x^2-10x+25=38$

${(x - 5)}^{2} = 38$ $(x-5)^2=38$

$x - 5 = \pm \sqrt{38}$ $x-5=+-sqrt38$

$x = 5 \pm \sqrt{38}$ $x=5+-sqrt38$

How do you solve x^2- 10x= 13x2−10x=13x^2- 10x= 13 by completing the square?