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

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Answer 1 · 2016-06-06T03:58:12

#x=1# or #x=-3#

According to Quadratic formula the solution of a quadratic equation #ax^2+bx+c=0# is

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

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

#hArr# #x^2+2x-3=0#, its solution is

#x=(-2+-sqrt(2^2-4xx1xx(-3)))/(2xx1)#

= #(-2+-sqrt(4+12))/2=(-2+-sqrt16)/2=(-2+-4)/2#

Hence #x=1# or #x=-3#