$(\dfrac{6}{7}+2x)^2=\dfrac{36}{25}$
$\leftrightarrow (\dfrac{6}{7}+2x)^2=(\dfrac{4}{5})^2$
$\leftrightarrow \left[ \begin{array}{l}\dfrac{6}{7}+2x=\dfrac{4}{5}\\\dfrac{6}{7}+2x=-\dfrac{4}{5}\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}2x=\dfrac{4}{5}-\dfrac{6}{7}\\2x=-\dfrac{4}{5}-\dfrac{6}{7}\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}2x=-\dfrac{2}{35}\\2x=-\dfrac{58}{35}\end{array}\right.$
$\leftrightarrow \left[ \begin{array}{l}x=-\dfrac{1}{35}\\x=-\dfrac{29}{35}\end{array} \right.$
Vậy `x=-1/35, x=-29/35`
OR: `S={-1/35;-29/35}`
