How do you solve # 2x(x - 1) = 3#?

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Answer 1 · 2016-03-26T06:08:14

#x=[1+-sqrt(13)]/4#

#2x(x-1)=3#

#4x^2-2x=3#

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

Comparing with #ax^2+bx+c=0# we get,

#a=4, b=-2,c=-3#

By Formula method,

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

#x=[-(-2)+-sqrt((-2)^2-4xx4xx-(3))]/(2xx4)#

#x=[2+-sqrt(4+48)]/(8)#

#x=[2+-sqrt(52)]/(8)#

#x=[2+-sqrt(4xx13)]/(8)#

#x=[2+-2sqrt(13)]/(8)#

#x=(2[1+-sqrt(13)])/8#

#x=[1+-sqrt(13)]/4#