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?

Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $\left(\frac32+\frac{\sqrt7}2,-\frac12+\frac{\sqrt7}2\right)$ and $\left(\frac32-\frac{\sqrt7}2,-\frac12-\frac{\sqrt7}2\right)$.
