How do you solve #x^2-x= -1# using the quadratic formula?

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Answer 1 · 2016-07-26T15:50:38

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

The quadratic formula for a general quadratic equation #ax^2+bx+c=0# is given by:

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

For the equation:

#x^2 - x = -1#

or #x^2 -x + 1=0#

you get

#a=1; b=-1 and c = 1#

by substituting these values in the quadratic formula:

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

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

or #x=(1+-sqrt(-3))/2#

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