\[\begin{array}{*{35}{l}}

{{x}^{2}}~+\text{ }1\text{ }=\text{ }0  \\

Since,\text{ }{{i}^{2}}~=\text{ }-1~\Rightarrow ~1\text{ }=\text{ }-{{i}^{2}}  \\

~substituting\text{ }1\text{ }=\text{ }-{{i}^{2}}  \\

{{x}^{2}}~-\text{ }{{i}^{2}}~=\text{ }0  \\

\left[ \text{ }{{a}^{2}}~-\text{ }{{b}^{2}}~=\text{ }\left( a\text{ }+\text{ }b \right)\text{ }\left( a\text{ }-\text{ }b \right) \right]  \\

\left( x\text{ }+\text{ }i \right)\text{ }\left( x\text{ }-\text{ }i \right)\text{ }=\text{ }0  \\

x\text{ }+\text{ }i\text{ }=\text{ }0\text{ }or\text{ }x\text{ }-\text{ }i\text{ }=\text{ }0  \\

x\text{ }=\text{ }-i\text{ }or\text{ }x\text{ }=\text{ }i  \\

\end{array}\]

∴ The roots of the given equation are i, -i