Find the roots of the following quadratic equation,if they exist,by the method of completing the square: $2x^2 - 7x + 3 = 0$.

(3, 1/2) Given equation: $2x^2 - 7x + 3 = 0$.
Step $1$: Move the constant term to the right side:
$2x^2 - 7x = -3$.
Step $2$: Divide the entire equation by the coefficient of $x^2$ (which is $2$):
$x^2 - \frac{7}{2}x = -\frac{3}{2}$.
Step $3$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-\frac{7}{2}$,so half of it is $-\frac{7}{4}$. Adding $(-\frac{7}{4})^2 = \frac{49}{16}$ to both sides:
$x^2 - 2(x)(\frac{7}{4}) + (\frac{7}{4})^2 = -\frac{3}{2} + \frac{49}{16}$.
Step $4$: Write the left side as a perfect square:
$(x - \frac{7}{4})^2 = \frac{-24 + 49}{16} = \frac{25}{16}$.
Step $5$: Take the square root of both sides:
$x - \frac{7}{4} = \pm \frac{5}{4}$.
Step $6$: Solve for $x$:
Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$.
Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$.
Thus,the roots are $3$ and $\frac{1}{2}$.