How do you solve #6-p^2/8=-4#?

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Answer 1 · 2016-10-30T09:41:26

#p=+-4sqrt5#

#6-p^2/8=-4#

#hArr6+4=p^2/8#

or #p^2/8=10#

or #p^2=80#

or #p^2-80=0#

or #p^2-(sqrt80)^2=0#

or #(p+sqrt80)(p-sqrt80)=0#

i.e. either #p+sqrt80=0# or #p-sqrt80=0#

i.e. either #p=-sqrt80# or #p=sqrt80#

i.e. #p=+-sqrt(4×4×5)=+-4sqrt5#