Multiply $ \dfrac{4a^2+a}{6a-4} $ by $ \dfrac{2a+10}{a^2-3a} $ to get $ \dfrac{4a^2+21a+5}{3a^2-11a+6} $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{4a^2+a}{6a-4} \cdot \frac{2a+10}{a^2-3a} & \xlongequal{\text{Step 1}} \frac{ \left( 4a+1 \right) \cdot \color{blue}{a} }{ \left( 3a-2 \right) \cdot \color{red}{2} } \cdot \frac{ \left( a+5 \right) \cdot \color{red}{2} }{ \left( a-3 \right) \cdot \color{blue}{a} } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 4a+1 }{ 3a-2 } \cdot \frac{ a+5 }{ a-3 } \xlongequal{\text{Step 3}} \frac{ \left( 4a+1 \right) \cdot \left( a+5 \right) }{ \left( 3a-2 \right) \cdot \left( a-3 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 4a^2+20a+a+5 }{ 3a^2-9a-2a+6 } = \frac{4a^2+21a+5}{3a^2-11a+6} \end{aligned} $$