Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

2x² - 6x + 3 = 0

Comparing 2x² - 6x + 3 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -6 and c = 3.

Substituting values in b² - 4ac, we get :

$\Rightarrow (-6)^2 - 4 \times 2 \times 3 \\[1em] \Rightarrow 36 - 24 \\[1em] \Rightarrow 12.$

Since, b² - 4ac > 0.

∴ There are two real and distinct roots.

Solving,

⇒ 2x² - 6x + 3 = 0

For distinct real roots,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 2 \times 3}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{6 \pm \sqrt{36 - 24}}{4} \\[1em] \Rightarrow x = \dfrac{6 \pm \sqrt{12}}{4} \\[1em] \Rightarrow x = \dfrac{6 \pm 2\sqrt{3}}{4} \\[1em] \Rightarrow x = \dfrac{3 \pm \sqrt{3}}{2}.$

Hence, x = $\dfrac{3 \pm \sqrt{3}}{2}.$