The synthetic division table is:
$$ \begin{array}{c|rrr}-1&10&5&13\\& & -10& \color{black}{5} \\ \hline &\color{blue}{10}&\color{blue}{-5}&\color{orangered}{18} \end{array} $$The solution is:
$$ \frac{ 10x^{2}+5x+13 }{ x+1 } = \color{blue}{10x-5} ~+~ \frac{ \color{red}{ 18 } }{ x+1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{-1}&10&5&13\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-1&\color{orangered}{ 10 }&5&13\\& & & \\ \hline &\color{orangered}{10}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrr}\color{blue}{-1}&10&5&13\\& & \color{blue}{-10} & \\ \hline &\color{blue}{10}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrr}-1&10&\color{orangered}{ 5 }&13\\& & \color{orangered}{-10} & \\ \hline &10&\color{orangered}{-5}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrr}\color{blue}{-1}&10&5&13\\& & -10& \color{blue}{5} \\ \hline &10&\color{blue}{-5}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 5 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrr}-1&10&5&\color{orangered}{ 13 }\\& & -10& \color{orangered}{5} \\ \hline &\color{blue}{10}&\color{blue}{-5}&\color{orangered}{18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x-5 } $ with a remainder of $ \color{red}{ 18 } $.