How do you solve #x^2= -2x + 3# using the quadratic formula?

1 Answer
Feb 27, 2017

Answer:

#x=-3,1#

Explanation:

#x^2=-2x+3#
subtract #x^2# from each side so you get 0 on the left
#0=-x^2-2x+3#
apply quadratic equation
#x=(-b\pm\sqrt(b^2-4ac))/(2a)\rArr(-(-2)\pm\sqrt((-2)^2-4(-1)(3)))/(2(-1))#
simplify and solve for #x#
#x=(2\pm\sqrt(4+12))/(-2)=(2\pm\sqrt(16))/(-2)=(2\pm4)/(-2)#
#(2+4)/-2\larrx\rarr(2-4)/-2#

therefore, #x# is either of the following...
#x=6/-2=-3#
#x=(-2)/-2=1#