Cách giải:
$3x^2+2x+5=0$
$→x^2+\dfrac{2}{3}x+\dfrac{5}{3}=0$
$→x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{14}{9}=0$
$→(x+\dfrac{1}{3})^2+\dfrac{14}{9}=0$
$→(x+\dfrac{1}{3})^2=-\dfrac{14}{9}$
Vì $(x+\dfrac{1}{3})^2 \geq 0$ mà $-\dfrac{14}{9}<0$
$→$pt vô nghiệm.