Solutions:

$\left( i \right)\text{ }Given,~{{x}^{2}}~\text{ }3x~\text{ }10\text{ }=0$

Taking LHS,

$=>{{x}^{2}}~\text{ }5x~+\text{ }2x~\text{ }10$

$=>x\left( x~\text{ }5 \right)~+\text{ }2\left( x~\text{ }5 \right)$

$=>\left( x~\text{ }5 \right)\left( x~+~2 \right)$

The roots of this equation, ${{x}^{2}}~\text{ }3x~\text{ }10\text{ }=\text{ }0$ are the values of x for which $\left( x~\text{ }5 \right)\left( x~+~2 \right)\text{ }=\text{ }0$

Therefore, $x~\text{ }5\text{ }=\text{ }0\text{ }or~x~+\text{ }2\text{ }=\text{ }0$

$=>~x~=\text{ }5\text{ }or~x~=\text{ }-2$

$\left( ii \right)\text{ }Given,\text{ }2{{x}^{2}}~+~x~\text{ }6\text{ }=\text{ }0$

Taking LHS,

$=>\text{ }2{{x}^{2}}~+\text{ }4x~\text{ }3x~\text{ }6$

$=>\text{ }2x\left( x~+~2 \right)\text{ }\text{ }3\left( x~+\text{ }2 \right)$

$=>\text{ }\left( x~+\text{ }2 \right)\left( 2x~\text{ }3 \right)$

The roots of this equation, $2{{x}^{2}}~+~x~\text{ }6=0$  are the values of x for which $\left( x~\text{ }5 \right)\left( x~+~2 \right)\text{ }=\text{ }0$

Therefore, $x~+\text{ }2~=\text{ }0\text{ }or~2x~\text{ }3\text{ }=\text{ }0$

$=>~x~=\text{ }-2\text{ }or~x~=\text{ }3/2$