As we have to solve the quadratic equation by completing square method and #2x^2# is not a square, let us first divide #2x^2+7x-4=0# by #2#, which gives us

#x^2+7/2x-2=0#. ....**(A)**

Now recall the identity #(x+a)^2=x^2+2ax+a^2# which shows that to complete square we must add square of half of #x's# coefficient. As coefficient of #x# is #7/2#, we add and subtract #(7/4)^2=49/16# and equation (A) becomes

#x^2+7/2x+49/16-49/16-2=0# or

#(x+7/4)^2-(49+32)/16=0# or

#(x+7/4)^2-81/16=0# or

#(x+7/4)^2-(9/4)^2=0# and using identity #(x^2-a^2)=(x+a)(x-a)#, we get

#(x+7/4+9/4)(x+7/4-9/4)=0# or

#(x+16/4)(x-2/4)=0#

#(x+4)(x-1/2)=0#

i.e. #x=-4# or #x=1/2#