Find the nature of the roots of the following quadratic equation. If the real roots exist,find them:
$2x^2 - 6x + 3 = 0$

(A) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
$(A)$ If $D > 0$,there are two distinct real roots.
$(B)$ If $D = 0$,there are two equal real roots.
$(C)$ If $D < 0$,there are no real roots.
Given the equation $2x^2 - 6x + 3 = 0$,comparing it with $ax^2 + bx + c = 0$,we get:
$a = 2, b = -6, c = 3$.
Discriminant $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$.
Since $D > 0$,the equation has two distinct real roots.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4}$.
Dividing the numerator and denominator by $2$,we get:
$x = \frac{3 \pm \sqrt{3}}{2}$.
Thus,the roots are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.