\(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}\cdot\left(0,2+\dfrac{12}{25}\right)=\dfrac{1}{5}+1\dfrac{6}{25}\)
=>\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{5}+\dfrac{31}{25}-\dfrac{17}{25}\left(\dfrac{1}{5}+\dfrac{12}{25}\right)\)
=>\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{36}{25}-\dfrac{17}{25}\cdot\dfrac{17}{25}=\left(\dfrac{6}{5}-\dfrac{17}{25}\right)^2=\left(\dfrac{13}{25}\right)^2\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{13}{25}\\x+\dfrac{1}{5}=-\dfrac{13}{25}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{25}-\dfrac{1}{5}=\dfrac{8}{25}\\x=-\dfrac{13}{25}-\dfrac{1}{5}=-\dfrac{18}{25}\end{matrix}\right.\)
