How do you complete the square to solve `-x^2-2x+3= 0`?

If you have to solve the equation you can multiply all by -1:
`−x^2−2x+3=0 -> x^2+2x-3=0`
It's always more simple to work with positive numbers =)
`x^2+2x-3=0` is similar to a perfect square trinomial that is `x^2+2x+1=(x+1)^2`.
So we can write our equation adding (or subtracting) what is missing to obtain a perfect square:
`x^2+2x-3=(x+1)^2-4=0`
If you want to solve the equation you have to obtain 4 in the square `(x+1)^2=4`, so `x+1=+-2` and the solutions are:
`x=1 or x=-3`

A perfect square trinomial is in the form `(a+b)^2=a^2+2ab+b^2`.

`-x^2-2x+3=0`

Multiply the equation times `-1`.

`x^2+2x-3=0`

The left side of the equation can be made into a perfect square trinomial by completing the square. This will enable the equation to be solved for `x`.

Add `3` to both sides.

`x^2+2x=3`

Divide the coefficient of the `x` term by `2`, then square the result. Add it to both sides of the equation.

`(2/2)^2=1`

`x^2+2x+1=4`

There is now a perfect square trinomial on the left side of the equation.

`(a+b)^2=a^2+2ab+b^2`

`a=x`
`b=1`

`(x+1)^2=4`

Take the square root of both sides.

`x+1=+-sqrt4`

`x+1=+-2`

If `x+1=2`, `x=1`.

If `x+1=-2`, `x=-3`