(i) 2x^{2} – 3x + 5 = 0

(ii) 3x^{2} – 4√3x + 4 = 0

(iii) 2x^{2} – 6x + 3 = 0

**Solution**

**(i)** *2x*^{2} – 3*x* + 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} – 4*ac*

*= *( – 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} – 3*x* + 5 = 0

**(ii) 3 x^{2} – 4√3x + 4 = 0**

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

*a* = 3, *b* = -4√3 and *c* = 4

We know, Discriminant = *b*^{2} – 4*ac*

= (-4√3)^{2 }– 4(3)(4)

= 48 – 48 = 0

As *b*^{2} – 4*ac* = 0,

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

Hence, the roots will be –*b*/2*a* and –*b*/2*a*.

–*b*/2*a *= -(-4√3)/2 × 3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

**(iii) 2 x^{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} – 4*ac*

= (-6)^{2} – 4 (2) (3)

= 36 – 24 = 12

As *b*^{2} – 4*ac* > 0,

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

x = [-b ± √(b^{2} – 4ac)]/2a

= (-(-6) ± √(-6^{2}-4(2)(3)) )/ 2(2)

= (6 ± 2√3 )/4

= (3 ± √3)/2

Therefore, the roots for the given equation are (3 + √3)/2 and (3 – √3)/2