Solution:

(i) Given,$2x^{2} – 3x + 5$ = 0

Comparing the equation with $ax^{2} + bx + c $= 0, we get

a = 2, b = -3 and c = 5

We know, discriminant = $b^{2} – 4ac$

= $( – 3)^{2}– 4 (2) (5)$ = 9 – 40

= – 31

As you can see, $b^{2} – 4ac$ < 0

Therefore, no real root is possible for the given equation, $2x^{2}$ – 3x + 5 = 0.

(ii) $3x^{2} – 4\sqrt{3}x$ + 4 = 0

Comparing the equation with $ax^{2}$ + bx + c = 0, we get

a = 3, b = $-4\sqrt{3}$ and c = 4

We know, Discriminant = $b^{2}$ – 4ac

= $(-4\sqrt{3})^{2}$ – 4(3)(4)

= 48 – 48 = 0

As $b^{2} – 4ac$ = 0,

Real roots exist for the given equation, and they are equal to each other.

Hence, the roots will be $\dfrac{–b}{2a}$ and $\dfrac{–b}{2a}$.

$\dfrac{–b}{2a}$

=$ \dfrac{-(-4\sqrt{3})}{2×3}$

= $\dfrac{4\sqrt{3}}{6} $

= $\dfrac{2\sqrt{3}}{3}$

= $\dfrac{2}{\sqrt{3}}$

Therefore, the roots are $\dfrac{2}{\sqrt{3}}$ and $\dfrac{2}{\sqrt{3}}$

(iii) $2x^{2} – 6x + 3$ = 0

Comparing the equation with $ax^{2} + bx + c$ = 0, we get

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

As we know, discriminant = $b^{2} – 4ac$

= $(-6)^{2}$ – 4 (2) (3)

= 36 – 24 = 12

As$ b^{2} – 4ac$ > 0,

Therefore, there are distinct real roots that exist for this equation, $2x^{2} – 6x + 3$ = 0.

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

= $\dfrac{(-(-6) ± \sqrt{(-62-4(2)(3)) )}}{ 2(2)}$

=$ \dfrac{(6±2\sqrt{3} )}{4}$

=$\dfrac{(3±\sqrt{3})}{2}$

Therefore, the roots for the given equation are $\dfrac{(3+\sqrt{3})}{2}$and $\dfrac{(3-\sqrt{3})}{2}$