$\left( i \right)\text{ }2{{x}^{2}}~~6x~+\text{ }3\text{ }=\text{ }0$

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

a = 2, b = -6, c = 3

a = 2, b = -6, c = 3

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

$=\text{ }{{\left( -6 \right)}^{2}}~\text{ }4\text{ }\left( 2 \right)\text{ }\left( 3 \right)$

$=\text{ }36\text{ }\text{ }24\text{ }=\text{ }12$

$=\text{ }36\text{ }\text{ }24\text{ }=\text{ }12$

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

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

$=\text{ }(-\left( -6 \right)\text{ }\pm \text{ }\surd (-{{6}^{2}}-4\left( 2 \right)\left( 3 \right))\text{ })/\text{ }2\left( 2 \right)$

$=\text{ }\left( 6\pm 2\surd 3\text{ } \right)/4$

$=\text{ }\left( 3\pm \surd 3 \right)/2$

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