How do you solve #2x^2+3x=20# by completing the square?

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Answer 1 · 2016-06-04T20:00:12

Factor out the two and cancel it from both sides to get
#x^2+3/2x=10#

We have

#2x^2+3x=20#

We factor out
#2(x^2+3/2x)=20#

#x^2+3/2x=10#

Divide split the linear term up so we have #b=2*(b/2)#
#x^2+2*(3/4)x=10#

#x^2+2*(3/4)x+3^2/4^2=10+3^2/4^2#

#x^2+2*(3/4)x+3^2/4^2=10+9/16#

#x^2+2*(3/4)x+3^2/4^2=160/16+9/16#

#(x+3/4)^2=169/16#

#(x+3/4)=pm 13/4#

#x+3/4-3/4=pm 13/4-3/4#

#x=-3/4 pm 13/4#