TITLE: Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$
QUESTION [5 upvotes]: Find all  points of the form $(x, −1)$, which are $4$ units from the point 
  $(3, 2)$.

I understand the distance formula, I think.  Also, I don't know how to format for math on here yet so I apologize for that.
\begin{align}
&\text{dist} = 4 = \sqrt{(x-3)^2 + (2-(-1))^2)} \\
&\implies 16 = (x-3)^2 + (2 + 1)^2 \tag{square both sides} \\
&\implies 16 = x^2 -6x + 18 \tag{expand} \\
&\implies 0 = x^2 -6x +2 \tag{zero on the left}
\end{align}
Next should be finding the factors which would give me the answers but I have no idea what would work for this.  
$$(x + \text{something})(x - \text{something}).$$
I must be doing something wrong or missing something.  Please correct me.

REPLY [3 votes]: You can work this out geometrically.  Plot the point $(3,2)$ and the horizontal line $y=-1$.  You want to find the two points on the the line which are $4$ units away from $(3,2)$.  Just mark about where they are.  Then there are two right triangles with vertices $(3,2)$, $(3,-1)$ and either of the points you marked.  The hypotenuse of each triangle is $4$.  The vertical leg is $3$ so the length of the other leg is $\sqrt{4^2-3^2} = \sqrt{7}.$  If you move $\sqrt{7}$ units left and right of $(3,-1)$, you get your two points $(3 \pm \sqrt{7},-1).$