\(pt\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2=\left(-\frac{2}{5}\right)^2\)
TH1: \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\Rightarrow x+\frac{1}{2}=\frac{2}{5}\Rightarrow x=\frac{-1}{10}\)
TH2: \(\left(x+\frac{1}{2}\right)^2=\left(-\frac{2}{5}\right)^2\Rightarrow x+\frac{1}{2}=\frac{-2}{5}\Rightarrow x=\frac{-9}{10}\)
Vậy \(x=\frac{-1}{10};\frac{-9}{10}\)