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?
 A: Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $\frac32\pm\frac{\sqrt 7}2$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(\frac32\pm\frac{\sqrt 7}2,-\frac12\pm\frac{\sqrt 7}2)$
A: Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\dfrac32\pm\dfrac12\sqrt{7},$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
A: 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 $y.$
$y = 1 - \frac {3+\sqrt 7}{2} = \frac {-1-\sqrt 7}{2}\\y = 1 - \frac {3-\sqrt 7}{2} = \frac {-1+\sqrt 7}{2}$
A: 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 $\left(\frac32-\frac{\sqrt7}2,-\frac12-\frac{\sqrt7}2\right)$.
A: After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$ 
$$ 2y^2+2y-3=0$$
which supplies corresponding roots 
$$ (y_1,y_2)= \frac{-1\pm \sqrt7}{2}.$$
