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How do you solve #x^2-7x+5=0# by completing the square?

1 Answer

Aug 7, 2016

#x=(7+-sqrt29)/2#

Explanation:

#x^2-7x+5=0#

#=>x^2-2*x*7/2+(7/2)^2-(7/2)^2+5=0#

#=>(x-7/2)^2=49/4-5=29/4#

#=>(x-7/2)=+-sqrt29/2#

#=>x=(7+-sqrt29)/2#

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