How do you solve #-3x^2-12=14x# by completing the square?

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Answer 1 · 2017-06-23T11:04:25

#x~~ -1.13 or x~~ -3.54#

#-3x^2-12=14x or 3x^2+14x+12 =0 or 3(x^2+14/3x)+12# or

# 3(x^2+14/3x +(7/3)^2)-(3*49/9)+12 =0#

#3(x+7/3)^2-49/3+12 =0 # or

#3(x+7/3)^2-13/3 =0 or 3(x+7/3)^2=13/3 # or

#(x+7/3)^2=13/9 or (x+7/3) = +-sqrt(13)/3#. Either

#x= -7/3+sqrt13/3 or x= -7/3-sqrt13/3# or

#x= (sqrt13-7)/3 or x= -(sqrt13+7)/3# or

#x~~ -1.13(2dp) or x~~ -3.54(2dp)# [Ans]