The synthetic division table is:
$$ \begin{array}{c|rrr}5&2&4&-59\\& & 10& \color{black}{70} \\ \hline &\color{blue}{2}&\color{blue}{14}&\color{orangered}{11} \end{array} $$The solution is:
$$ \frac{ 2x^{2}+4x-59 }{ x-5 } = \color{blue}{2x+14} ~+~ \frac{ \color{red}{ 11 } }{ x-5 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&4&-59\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 2 }&4&-59\\& & & \\ \hline &\color{orangered}{2}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&4&-59\\& & \color{blue}{10} & \\ \hline &\color{blue}{2}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 10 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrr}5&2&\color{orangered}{ 4 }&-59\\& & \color{orangered}{10} & \\ \hline &2&\color{orangered}{14}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 14 } = \color{blue}{ 70 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&2&4&-59\\& & 10& \color{blue}{70} \\ \hline &2&\color{blue}{14}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -59 } + \color{orangered}{ 70 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrr}5&2&4&\color{orangered}{ -59 }\\& & 10& \color{orangered}{70} \\ \hline &\color{blue}{2}&\color{blue}{14}&\color{orangered}{11} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x+14 } $ with a remainder of $ \color{red}{ 11 } $.