How to describe $G/U$? Let $G=SL_2(\mathbb{C})$ and let $U = \{\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C}\}$. We have an action of $U$ on $G$ by right multiplication. By definition, $G/U$ is the set of all $U$-orbits under this action. How to describe $G/U$ as a set?
Take $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$. The orbit of $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$ is the set $$ \{ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C} \}. $$
But I don't know how to classify orbits in $G/U$. Thank you very much.
Edit: $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$ and  $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right) \in G$ are in the same orbit. Indeed, Suppose that
 $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)$ =  $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right)$. Then $x = (b'-b)/a$. We also have $cx+d=(cb'+ad-bc)/a=(cb'+1)/a$. Since $ad'-b'c=1$, $(cb'+1)/a=d'$. Therefore $x = (b'-b)/a$ is a solution of $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)$ =  $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right)$. Hence $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$ and  $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right) \in G$ are in the same orbit.
I think that $G/U = \{ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right): a, c \in \mathbb{C}, b,d \text{ are fixed }\}$. Is this correct? Thank you very much.

Consider the tautological action of $G:= SL_2(\mathbb C)$ on $\mathbb C^2$.  It is transitive on the
points of $\mathbb C^2 \setminus \{0\}$, and the stabilizer of the vector $(1,0)$ is precisely $U$.  
So the quotient $G/U$ is naturally identified with $\mathbb C^2\setminus \{0\}$.
As user165670 notes in their answer, the map is given explicitly by sending a matrix
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ to the vector $(a,c)$ (since this
is the image of $(1,0)$ under the action of this matrix).
