How do you solve #(3x-2)^2-7=0#?

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Answer 1 · 2015-10-03T21:03:13

This equation has 2 solutions:

#x_1=(2-sqrt(7))/3#

#x_2=(2+sqrt(7))/3#

#(3x-2)^2-7=0#

#9x^2-12x+4-7=0#

#9x^2-12x-3=0#

#3x^2-4x-1=0#

#Delta=(-4)^2-4*3*(-1)#

#Delta=16+12#

#Delta=28#

#sqrt(Delta)=2sqrt(7)#

#x_1=(4-2sqrt(7))/6#

#x_1=(2-sqrt(7))/3#

#x_2=(2+sqrt(7))/3#