Find $ \left(x-1\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(x-1\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 1 + \color{red}{1^2} = x^2-2x+1\end{aligned} $$

Subtract $2x$ from $ \dfrac{x^2-2x+1}{2} $ to get $ \dfrac{ \color{purple}{ x^2-6x+1 } }{ 2 }$.

Step 1: Write $ 2x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{2}$.

$$ \begin{aligned} \frac{x^2-2x+1}{2} -2x & \xlongequal{\text{Step 1}} \frac{x^2-2x+1}{2} - \frac{2x}{\color{red}{1}} = \frac{ x^2-2x+1 }{ 2 } - \frac{ 2x \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^2-2x+1 } }{ 2 } - \frac{ \color{purple}{ 4x } }{ 2 }=\frac{ \color{purple}{ x^2-6x+1 } }{ 2 } \end{aligned} $$