How do you solve the equation #2x^2+3x-5=0# by completing the square?

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Answer 1 · 2017-02-03T11:22:33

The solutions are #S={1,-5/2}#

Let's complete the square

#2x^2+3x-5=0#

#2(x^2+3/2x)-5=0#

#2(x^2+3/2x+9/16)-5-9/8=0#

#2(x+3/4)^2-49/8=0#

#2(x+3/4)^2=49/16#

#(x+3/4)^2=49/16#

Taking the square roots

#(x+3/4)=+-7/4#

#x=-3/4+-7/4#

#x_1=-10/4=-5/2#

#x_2=4/4=1#