Add $3$ and $ \dfrac{3}{2q^2-24q+17} $ to get $ \dfrac{ \color{purple}{ 6q^2-72q+54 } }{ 2q^2-24q+17 }$.

Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2q^2-24q+17 }$.

$$ \begin{aligned} 3+ \frac{3}{2q^2-24q+17} & \xlongequal{\text{Step 1}} \frac{3}{\color{red}{1}} + \frac{3}{2q^2-24q+17} = \frac{ 3 \cdot \color{blue}{ \left( 2q^2-24q+17 \right) }}{ 1 \cdot \color{blue}{ \left( 2q^2-24q+17 \right) }} + \frac{ 3 }{ 2q^2-24q+17 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 6q^2-72q+51 } }{ 2q^2-24q+17 } + \frac{ \color{purple}{ 3 } }{ 2q^2-24q+17 }=\frac{ \color{purple}{ 6q^2-72q+54 } }{ 2q^2-24q+17 } \end{aligned} $$