Simplify denominator

$$ y^2+ \color{blue}{6y} + \color{blue}{8y} = y^2+ \color{blue}{14y} $$

Subtract $ \dfrac{2}{y^2+14y} $ from $ \dfrac{3}{y^2+y+12} $ to get $ \dfrac{ \color{purple}{ y^2+40y-24 } }{ y^4+15y^3+26y^2+168y }$.

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}{ y^2+14y }$ and the second by $\color{blue}{ y^2+y+12 }$.

$$ \begin{aligned} \frac{3}{y^2+y+12} - \frac{2}{y^2+14y} & = \frac{ 3 \cdot \color{blue}{ \left( y^2+14y \right) }}{ \left( y^2+y+12 \right) \cdot \color{blue}{ \left( y^2+14y \right) }} - \frac{ 2 \cdot \color{blue}{ \left( y^2+y+12 \right) }}{ \left( y^2+14y \right) \cdot \color{blue}{ \left( y^2+y+12 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3y^2+42y } }{ y^4+14y^3+y^3+14y^2+12y^2+168y } - \frac{ \color{purple}{ 2y^2+2y+24 } }{ y^4+14y^3+y^3+14y^2+12y^2+168y } = \\[1ex] &=\frac{ \color{purple}{ 3y^2+42y - \left( 2y^2+2y+24 \right) } }{ y^4+15y^3+26y^2+168y }=\frac{ \color{purple}{ y^2+40y-24 } }{ y^4+15y^3+26y^2+168y } \end{aligned} $$