Subtract $q-6$ from $ \dfrac{q}{3q+8} $ to get $ \dfrac{ \color{purple}{ -3q^2+11q+48 } }{ 3q+8 }$.

Step 1: Write $ q-6 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{3q+8}$.

$$ \begin{aligned} \frac{q}{3q+8} -q-6 & \xlongequal{\text{Step 1}} \frac{q}{3q+8} - \frac{q-6}{\color{red}{1}} = \frac{ q }{ 3q+8 } - \frac{ \left( q-6 \right) \cdot \color{blue}{ \left( 3q+8 \right) }}{ 1 \cdot \color{blue}{ \left( 3q+8 \right) }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ q } }{ 3q+8 } - \frac{ \color{purple}{ 3q^2+8q-18q-48 } }{ 3q+8 }=\frac{ \color{purple}{ q - \left( 3q^2+8q-18q-48 \right) } }{ 3q+8 } = \\[1ex] &=\frac{ \color{purple}{ -3q^2+11q+48 } }{ 3q+8 } \end{aligned} $$