$f(x)=2x^2-3x+5$
$=2(x^2-\dfrac{3}{2}x+\dfrac{5}{2})$
$=2(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{31}{16})$
$=2(x-\dfrac{3}{4})^2+\dfrac{31}{8}>0∀x$
Vì $(x-\dfrac{3}{4})^2≥0∀x⇒2(x-\dfrac{3}{4})^2≥0∀x⇒2(x-\dfrac{3}{4})^2+\dfrac{31}{8}≥\dfrac{31}{8}>0$
$⇒f(x)$ vô nghiệm