Solve the following equation:
\[{{x}^{2}}+2=0\]

Hint: We will just equate the equation to zero and take the number ‘2’ to the right hand side of equality and take the square root of both sides. We will get an imaginary root of the given equation.

Complete step-by-step answer:
We have been given the equation \[{{x}^{2}}+2=0\] is in the form of \[a{{x}^{2}}+bx+c=0\] where a = 1, b = 0 and c = 2.
Now we will calculate the value of discriminant (D) given by as follows:
\[D={{b}^{2}}-4ac\]
We have a = 1, b = 0 and c = 2.
\[\begin{align}
  & D={{(0)}^{2}}-4\times 1\times 2 \\
 & D=-8 \\
\end{align}\]
Here, the value of the discriminant (D) is a negative value and we know that if the value of (D) is negative then the equation has no real roots. It has only imaginary roots.
Now let us consider the given equation \[{{x}^{2}}+2=0\].
On subtracting 2 from both sides of the equation, we get,
\[\begin{align}
  & {{x}^{2}}+2-2=0-2 \\
 & {{x}^{2}}=-2 \\
\end{align}\]
After taking square root to both sides of the equation, we get,
\[\sqrt{{{x}^{2}}}=\sqrt{-2}\]
Hence \[x=\sqrt{2}i\] and \[x=-\sqrt{2}i\].
Since we know that the value of \[\sqrt{-1}\] is equal to iota \[(i)\].
Therefore, on solving the equation \[{{x}^{2}}+2=0\], we get the roots as \[\sqrt{2}i\] and \[-\sqrt{2}i\].

Note: Be careful while solving the equation and taking square root on both the sides of the equality. Also, remember that the value of \[\sqrt{-1}\] is known as iota and imaginary roots are always in a pair which means that magnitude will be the same but the algebraic signs will differ.