How do you solve # x^2-x= -1#?

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Answer 1 · 2015-07-28T19:28:36

#x=1/2+-isqrt(3)/2#

Complete the square:
#x^2-x=(x-1/2)^2-1/4=-1#

Rearrange to make #x# the subject:
#(x-1/2)^2-1/4=-1#

=>

#(x-1/2)^2=-3/4#

=>

#x-1/2=+-isqrt(3)/2#

=>

#x=1/2+-isqrt(3)/2#