The synthetic division table is:
$$ \begin{array}{c|rrr}-3&4&5&-14\\& & -12& \color{black}{21} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 4x^{2}+5x-14 }{ x+3 } = \color{blue}{4x-7} ~+~ \frac{ \color{red}{ 7 } }{ x+3 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{-3}&4&5&-14\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-3&\color{orangered}{ 4 }&5&-14\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&4&5&-14\\& & \color{blue}{-12} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrr}-3&4&\color{orangered}{ 5 }&-14\\& & \color{orangered}{-12} & \\ \hline &4&\color{orangered}{-7}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&4&5&-14\\& & -12& \color{blue}{21} \\ \hline &4&\color{blue}{-7}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 21 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrr}-3&4&5&\color{orangered}{ -14 }\\& & -12& \color{orangered}{21} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x-7 } $ with a remainder of $ \color{red}{ 7 } $.