find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one.  I'm teaching my self pre-calc out of a book and it isn't showing me how to do this.  I've been all over the internet and could only find a few examples.  I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in.  What am I doing wrong?
1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values.  These points are not on the line.  Can someone tell me what I'm doing wrong please?

You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say 
$x = \frac {3\pm\sqrt 7}{2}$ is simpler than $x = 1.5 \pm \sqrt {1.75}$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - \frac {3+\sqrt 7}{2} = \frac {-1-\sqrt 7}{2}\\y = 1 - \frac {3-\sqrt 7}{2} = \frac {-1+\sqrt 7}{2}$
