Find the roots of the quadratic equation by applying the quadratic formula:
$2 x^{2}-7 x+3=0$

(N/A) Given equation: $2 x^{2}-7 x+3=0$
On comparing this equation with the standard form $a x^{2}+b x+c=0$,we get:
$a=2, b=-7, c=3$
The quadratic formula is given by:
$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Substituting the values of $a, b,$ and $c$:
$x=\frac{-(-7) \pm \sqrt{(-7)^{2}-4(2)(3)}}{2(2)}$
$x=\frac{7 \pm \sqrt{49-24}}{4}$
$x=\frac{7 \pm \sqrt{25}}{4}$
$x=\frac{7 \pm 5}{4}$
Now,solving for both cases:
Case $1$: $x = \frac{7+5}{4} = \frac{12}{4} = 3$
Case $2$: $x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$
Therefore,the roots of the equation are $3$ and $\frac{1}{2}$.