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, when (i) x ∈ N (ii) x ∈ Q

Let us factorize the expression,

    \[\begin{array}{*{35}{l}} 2x{}^\text{2}\text{ }\text{ }9x\text{ }+\text{ }10\text{ }=\text{ }0  \\ 2{{x}^{2}}~\text{ }4x\text{ }\text{ }5x\text{ }+\text{ }10\text{ }=\text{ }0  \\ 2x\left( x\text{ }\text{ }2 \right)\text{ }\text{ }5\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0  \\ \left( 2x\text{ }\text{ }5 \right)\text{ }\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0  \\ \end{array}\]

So now,

   

(i) When, x ∈ N then, x =

   

(ii) When, x ∈ Q then, x =