The synthetic division table is:
$$ \begin{array}{c|rr}2&2&3\\& & \color{black}{4} \\ \hline &\color{blue}{2}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 2x+3 }{ x-2 } = \color{blue}{2} ~+~ \frac{ \color{red}{ 7 } }{ x-2 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rr}\color{blue}{2}&2&3\\& & \\ \hline && \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rr}2&\color{orangered}{ 2 }&3\\& & \\ \hline &\color{orangered}{2}& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rr}\color{blue}{2}&2&3\\& & \color{blue}{4} \\ \hline &\color{blue}{2}& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 4 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rr}2&2&\color{orangered}{ 3 }\\& & \color{orangered}{4} \\ \hline &\color{blue}{2}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2 } $ with a remainder of $ \color{red}{ 7 } $.