<s>[INST] Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$ 
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.
 [/INST] Suppose you have two points in a 2-D coordinate system such as $(x_{1}, y_{1}) = (3, 2)$ and $(x_{2}, y_{2}) = (x, -1)$ and distance between them is 4 units. Hence you can write as 
\begin{align} 
& d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} \\
\implies & 4 = \sqrt{(3-x)^{2} + (2-(-1))^{2}} \\
\implies & 16 = 9 + x^{2} - 6x +9 \tag{squaring both sides}\\
\implies & x^{2} - 6x + 2 = 0
\end{align}
This is a quadratic equation. The solution to a quadratic equation of the form $ax^{2} + bx + c$ can be found as $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$. Hence comparing the standard form and the equation above, we have $a = 1, b = -6$ and $c = 2$. Hence the roots will be 
$$x = \dfrac{6 \pm \sqrt{36 - 8}}{2} = \dfrac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}$$
</s>