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

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Answer 1 · 2016-09-11T18:01:12

$ax^2+bx+1/2b^2$

$x^2-10x=13, b=10, 1/2b^2=5^2=25$

Add 25 to both sides

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

$(x-5)^2=38$

$x-5=+-sqrt38$

$x=5+-sqrt38$

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