The synthetic division table is:
$$ \begin{array}{c|rrr}3&-2&-5&10\\& & -6& \color{black}{-33} \\ \hline &\color{blue}{-2}&\color{blue}{-11}&\color{orangered}{-23} \end{array} $$The solution is:
$$ \frac{ -2x^{2}-5x+10 }{ x-3 } = \color{blue}{-2x-11} \color{red}{~-~} \frac{ \color{red}{ 23 } }{ x-3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&-2&-5&10\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ -2 }&-5&10\\& & & \\ \hline &\color{orangered}{-2}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&-2&-5&10\\& & \color{blue}{-6} & \\ \hline &\color{blue}{-2}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrr}3&-2&\color{orangered}{ -5 }&10\\& & \color{orangered}{-6} & \\ \hline &-2&\color{orangered}{-11}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ -33 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&-2&-5&10\\& & -6& \color{blue}{-33} \\ \hline &-2&\color{blue}{-11}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -33 \right) } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrr}3&-2&-5&\color{orangered}{ 10 }\\& & -6& \color{orangered}{-33} \\ \hline &\color{blue}{-2}&\color{blue}{-11}&\color{orangered}{-23} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x-11 } $ with a remainder of $ \color{red}{ -23 } $.