$\dfrac{2}{5}x²+\dfrac{1}{3}x+\dfrac{1}{15}=0$
$Δ=(\dfrac{1}{3})²-4.\dfrac{2}{5}.\dfrac{1}{15}$
$=\dfrac{1}{9}-\dfrac{8}{75}$
$=\dfrac{1}{225}>0$
$→$ Pt có 2 nghiệm
\(\to\left[ \begin{array}{l}x_1=\dfrac{-\dfrac{1}{3}+\sqrt\Delta}{2.\dfrac{2}{5}}=\dfrac{\dfrac{-4}{15}}{\dfrac{4}{5}}=-\dfrac{1}{3}\\x_2=\dfrac{-\dfrac{1}{3}-\sqrt\Delta}{2.\dfrac{2}{5}}=\dfrac{-\dfrac{2}{5}}{\dfrac{4}{5}}=-\dfrac{1}{2}\end{array} \right.\)
Vậy pt có tập nghiệm $S=\bigg\{-\dfrac{1}{3};-\dfrac{1}{2}\bigg\}$
