How do you solve #x^2-x-1=0# using the quadratic formula?

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Answer 1 · 2017-07-24T07:22:41

See below

#x=1.62 or x= -0.62#

The quadratic formula is #[-b+-sqrt(b^2-4ac)]/(2a)#

this is used for formulas #ax^2+bx+c=0#

in your problem, #" "x^2−x−1=0" "# a=1, b=-1, c=-1#

substitute into quadratic #" "[-(-1)+-sqrt((-1)^2-4(1)(-1))]/(2*1)#

this equals #" "(1+-sqrt(1+4))/2#

#(1+sqrt5)/2 or (1-sqrt5)/2#

#x=1.62 or x= -0.62#