How do you solve #2x²-(5/2)x+2=0#?

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Answer 1 · 2017-07-02T07:58:44

#x=5/8+isqrt39/8# or #5/8-isqrt39/8#

#2x^2-5/2x+2=0# can be written as

#2(x^2-5/4 x+1)=0# - dividing by #2#

or #x^2-5/4 x+1=0#

or #x^2-2xx5/8xx x+(5/8)^2-(5/8)^2+1=0#

or #(x-5/8)^2-25/64+1=0#

or #(x-5/8)^2+39/64=0#

or #(x-5/8)^2-(-1xx39/64)=0#

or #(x-5/8)^2-(isqrt39/8)^2=0#

or #(x-5/8+isqrt39/8)(x-5/8-isqrt39/8)=0#

Hence #x=5/8+isqrt39/8# or #5/8-isqrt39/8#