Subtract $ \dfrac{7}{2} $ from $ \dfrac{3}{4v^2+4v} $ to get $ \dfrac{ \color{purple}{ -28v^2-28v+6 } }{ 8v^2+8v }$.

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}{ 2 }$ and the second by $\color{blue}{ 4v^2+4v }$.

$$ \begin{aligned} \frac{3}{4v^2+4v} - \frac{7}{2} & = \frac{ 3 \cdot \color{blue}{ 2 }}{ \left( 4v^2+4v \right) \cdot \color{blue}{ 2 }} - \frac{ 7 \cdot \color{blue}{ \left( 4v^2+4v \right) }}{ 2 \cdot \color{blue}{ \left( 4v^2+4v \right) }} = \\[1ex] &=\frac{ \color{purple}{ 6 } }{ 8v^2+8v } - \frac{ \color{purple}{ 28v^2+28v } }{ 8v^2+8v }=\frac{ \color{purple}{ 6 - \left( 28v^2+28v \right) } }{ 8v^2+8v } = \\[1ex] &=\frac{ \color{purple}{ -28v^2-28v+6 } }{ 8v^2+8v } \end{aligned} $$