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