#2x^2+8x-3=0#

#rArr color(blue)2(x^2+4x)=3#

If #color(orange)(4x)# is the middle term of a squared binomial with the form

#color(white)("XXX")(x+a)^2 = (x^2+color(orange)(2ax)+a^2)#

then

#color(white)("XXX")color(orange)a=color(orange)2#

and

#color(white)("XXX")color(green)(a^2)=color(green)(4)# must be added to #x^2+4x# to complete the square.

This will result in adding #color(blue)2 xxcolor(green)4# to the left side of the equation;

so we must add this to the right side as well to keep the equation valid.

#color(white)("XXX")color(blue)2(x^2+4x+color(green)4)=3+color(blue)2xxcolor(green)4#

Rewriting the parenthesized factor as a squared binomial and simplifying.

#color(white)("XXX")2(x+2)^2=11#

Now divide both sides by #2#, leaving only a squared binomial on the left:

#color(white)("XXX")(x+2)^2=11/2#

Take the square root of both sides (don't forget that there will be both a positive and negative root on the right side):

#color(white)("XXX")x+2=+-sqrt(11/2)#

Subtract #2# from both sides to isolate the variable #x#

#color(white)("XXX")x=-2+-sqrt(11/2)#