How do you solve `2x^2 + 8x -3 =0` by completing the square?

`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)`

`2x^2 +8x -3 = 0 or 2(x^2+4x) -3 =0 or 2(x^2+4x +4) -3 =8` [Adding 8 on both sides]

or `2(x+2)^2 =11 or (x+2)^2 = 11/2 or x+2 = +-sqrt(11/2)`

or`x = -2 +- sqrt(11/2) = -2 +- 1/2(sqrt22) :. x ~~0.345 or x ~~ -4.345` [Ans]