Subtract $ \dfrac{2}{4+i} $ from $ \dfrac{3}{2-i} $ to get $ \dfrac{ \color{purple}{ 5i+8 } }{ -i^2-2i+8 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ i+4 }$ and the second by $\color{blue}{ -i+2 }$.

$$ \begin{aligned} \frac{3}{2-i} - \frac{2}{4+i} & = \frac{ 3 \cdot \color{blue}{ \left( i+4 \right) }}{ \left( 2-i \right) \cdot \color{blue}{ \left( i+4 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( -i+2 \right) }}{ \left( 4+i \right) \cdot \color{blue}{ \left( -i+2 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3i+12 } }{ 2i+8-i^2-4i } - \frac{ \color{purple}{ -2i+4 } }{ 2i+8-i^2-4i }=\frac{ \color{purple}{ 3i+12 - \left( -2i+4 \right) } }{ -i^2-2i+8 } = \\[1ex] &=\frac{ \color{purple}{ 5i+8 } }{ -i^2-2i+8 } \end{aligned} $$

$$ -i^2 = -(-1) = 1 $$

Simplify denominator

$$ \color{blue}{1} -2i+ \color{blue}{8} = -2i+ \color{blue}{9} $$

Divide $ \, 8+5i \, $ by $ \, 9-2i \, $ to get $\,\, \dfrac{62+61i}{85} $.
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