In order to complete the square, since we have the term -10x, we need the term 5² = 25 to have a perfect square. So adding this term to both sides of the equation, we have:
[tex]\begin{gathered} x^2-10x+8=0 \\ x^2-10x+25+8=25 \\ (x-5)^2+8=25 \\ \frac{\mleft(2x-10\mright)^2}{4}=17_{} \\ \mleft(2x-10\mright)^2=68 \end{gathered}[/tex]
The left side of the equation is (2x - 10)², therefore the correct option is the third one.
