How do you solve the equation #2x^2-3x-3=0# by completing the square?

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Answer 1 · 2016-11-10T14:03:29

#x=(3+-sqrt(15))/2#

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

divide by #2# to make coefficient #x^2=1#

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

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

square half coefficient of #x # then add and subtract

#(x^2-(3x)/2+(3/4)^2)-3/2-(3/4)^2=0#

the first bracket is now a perfect square

#(x-3/2)^2-3/2-9/4=0#

#(x-3/2)^2-15/4=0#

now rearrange for #x#

#(x-3/2)^2=15/4#

#x-3/2=+-sqrt(15/4)#

#x-3/2=+-sqrt(15)/2#

#x=3/2+-sqrt(15)/2#

#x=(3+-sqrt(15))/2#