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Answer 1 · 2017-12-14T16:39:33

#x = {(1- sqrt(7)i)/2, (1+ sqrt(7)i)/2}#

#x^2-x+1=-1#

#=>x^2-x+2=0#

Using,

For, #ax^2+bx+c=0#

#x = (-b+- sqrt(b^2-4ac))/(2a)#

Here, #a=1#, #b=-1# and #c=2#.

#x = (-(-1)+- sqrt((-1)^2-4(1)(2)))/(2(1))#

#x = (1+- sqrt(1-8))/2#

#x = (1+- sqrt(7)i)/2#; where #i = sqrt(-1)#

So, we can see that both the roots of this equation are imaginary numbers.