The synthetic division table is:
$$ \begin{array}{c|rrr}5&2&-1&-2\\& & 10& \color{black}{45} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{orangered}{43} \end{array} $$The solution is:
$$ \frac{ 2x^{2}-x-2 }{ x-5 } = \color{blue}{2x+9} ~+~ \frac{ \color{red}{ 43 } }{ x-5 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&-1&-2\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 2 }&-1&-2\\& & & \\ \hline &\color{orangered}{2}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&-1&-2\\& & \color{blue}{10} & \\ \hline &\color{blue}{2}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 10 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrr}5&2&\color{orangered}{ -1 }&-2\\& & \color{orangered}{10} & \\ \hline &2&\color{orangered}{9}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 9 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&-1&-2\\& & 10& \color{blue}{45} \\ \hline &2&\color{blue}{9}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 45 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrr}5&2&-1&\color{orangered}{ -2 }\\& & 10& \color{orangered}{45} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{orangered}{43} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x+9 } $ with a remainder of $ \color{red}{ 43 } $.