How do you factor and solve #-x^2-2x+3 = 0#?

#(-x+1)(x+3)=0#

#-x+1=0 or x+3=0#

#x=1 or x=-3#

Given: #-x^2-2x+3=0#

First of all I am going to make a deliberate mistake.

Notice that #1xx3=3 and -3+1=-2#

So in error write: #(x-3)(x+1) color(red)(larr" Does not give "-x^2)#

Lets consider just the #x# and ignore the potential error about the constant.

#(-x-3)(x+1) = -x^2color(red)(-3x-x)-3#

Because we have a negative #x# lets use this to give us
the required #-3x+x->-2x#

#(-x+?)(x+?) rarr" we need "color(magenta)(-3x)# so do this:

#color(green)((-x+?)(xcolor(magenta)(+3)) = -x^2-3x+?x+(?xx3))#

We now need #+3# so #?xx3# has to be #(color(magenta)(+1))xx(+3)#

#color(green)((-xcolor(magenta)(+1))(x+3)color(white)("dd") =color(white)("dd") -x^2color(white)("d")ubrace(-3x+x)+3 =0#
#color(green)(color(white)("dddddddddddddddddddd")-x^2color(white)("dd")-2xcolor(white)("dd")+3=0)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the factorisation is #(-x+1)(x+3)=0#

Set #-x+1=0 => x=1#
Set #+x+3=0=>x=-3#

#-x^2 - 2x + 3 = 0#
Since a + b + c = 0, use shortcut.
The 2 real roots are:
x = 1 , and #x = c/a = 3/-1 = -3.#

Rule of Shortcut.
If a + b + c = 0, The 2 real roots are: x = 1, and #x = c/a#
If a - b + c = 0, The 2 real roots are: x = -1, and #x = -c/a#

Remember this shortcut. It will save you a lot of time and effort.