How do you solve #x^2+5=167#?

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Answer 1 · 2017-02-05T13:23:32

#x=+-9*sqrt2# or #x=+-12.728# to the nearest 3 decimals

#x^2+5=167" "# there is no #x# term change to #x^2 = c#

#:.x^2=162#

#:.x=+-sqrt162#

#:.x=+-(sqrt3*sqrt3*sqrt3*sqrt3*sqrt2)#

#sqrt3*sqrt3=3#

#:.x=+-(3*3*sqrt2)#

#:.x=+-9sqrt2#

Or #x=+-9*1.414213562#

#:.x=+-12.728# to the nearest 3 decimals.