Multiply $ \color{blue}{4} $ by $ \left( x+2\right) $ $$ \color{blue}{4} \cdot \left( x+2\right) = 4x+8 $$

Multiply $ \color{blue}{3} $ by $ \left( x+2\right) $ $$ \color{blue}{3} \cdot \left( x+2\right) = 3x+6 $$

Simplify $ \dfrac{x+2}{4x+8} $ to $ \dfrac{1}{4} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+2}$.

$$ \begin{aligned} \frac{x+2}{4x+8} & =\frac{ 1 \cdot \color{blue}{ \left( x+2 \right) }}{ 4 \cdot \color{blue}{ \left( x+2 \right) }} = \\[1ex] &= \frac{1}{4} \end{aligned} $$

Subtract $ \dfrac{2}{3x+6} $ from $ \dfrac{1}{4} $ to get $ \dfrac{ \color{purple}{ 3x-2 } }{ 12x+24 }$.

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}{ 3x+6 }$ and the second by $\color{blue}{ 4 }$.

$$ \begin{aligned} \frac{1}{4} - \frac{2}{3x+6} & = \frac{ 1 \cdot \color{blue}{ \left( 3x+6 \right) }}{ 4 \cdot \color{blue}{ \left( 3x+6 \right) }} - \frac{ 2 \cdot \color{blue}{ 4 }}{ \left( 3x+6 \right) \cdot \color{blue}{ 4 }} = \\[1ex] &=\frac{ \color{purple}{ 3x+6 } }{ 12x+24 } - \frac{ \color{purple}{ 8 } }{ 12x+24 }=\frac{ \color{purple}{ 3x-2 } }{ 12x+24 } \end{aligned} $$