(i) Given,

$2{{x}^{2}}~\text{ }3x~+\text{ }5\text{ }=\text{ }0$

Comparing the equation with $a{{x}^{2}}~+~bx~+~c~=\text{ }0$ , we get

a = 2, b = -3 and c = 5

We know, Discriminant $=~{{b}^{2}}~\text{ }4ac$

$=~{{\left( \text{ }\text{ }3 \right)}^{2}}~\text{ }4\text{ }\left( 2 \right)\text{ }\left( 5 \right)\text{ }=\text{ }9\text{ }\text{ }40$

$=\text{ }\text{ }31$

.As you can see, ${{b}^{2}}~\text{ }4ac\text{ }<\text{ }0$

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

(ii) $3{{x}^{2}}~\text{ }4\surd 3x~+\text{ }4\text{ }=\text{ }0$

Comparing the equation with $a{{x}^{2}}~+~bx~+~c~=\text{ }0$ , we get

a = 3, b = -4√3 and c = 4

We know, Discriminant $=~{{b}^{2}}~\text{ }4ac$

$=\text{ }{{\left( -4\surd 3 \right)}^{2~}}\text{ }4\left( 3 \right)\left( 4 \right)$

$=\text{ }48\text{ }\text{ }48\text{ }=\text{ }0$

$As~{{b}^{2}}~\text{ }4ac~=\text{ }0$ ,

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

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

$b/2a~=\text{ }-\left( -4\surd 3 \right)/2\times 3\text{ }=\text{ }4\surd 3/6\text{ }=\text{ }2\surd 3/3\text{ }=\text{ }2/\surd 3$

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