How do you solve ${{\left( x+2 \right)}^{2}}=-7$ ?

Hint: We have to find the solution of ${{\left( x+2 \right)}^{2}}=-7$. Then we take the square root on both sides of the equation. The right-hand side being negative, we get imaginary value as the square root value. From that we subtract 2 to the both sides to find the value of $x$ for ${{\left( x+2 \right)}^{2}}=-7$. Then we put the solution value in the equation to verify the result.

Complete step-by-step solution:
We need to find the solution of the given equation ${{\left( x+2 \right)}^{2}}=-7$.
We take square roots on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\left( x+2 \right)}^{2}}}=\sqrt{-7}=\pm i\sqrt{7} \\
 & \Rightarrow \left( x+2 \right)=\pm i\sqrt{7} \\
\end{align}$
Here $i$ is the complex value.
Now we subtract 2 to the both sides of the equation $\left( x+2 \right)=\pm i\sqrt{7}$ to get value for variable $x$.
$\begin{align}
  & \left( x+2 \right)-2=\pm i\sqrt{7}-2 \\
 & \Rightarrow x=-2\pm i\sqrt{7} \\
\end{align}$
The given quadratic equation has two solutions and they are $x=-2\pm i\sqrt{7}$.

Note: We try to verify the value of the root of $x=-2\pm i\sqrt{7}$ for the equation ${{\left( x+2 \right)}^{2}}=-7$.
Putting the value $x=-2+i\sqrt{7}$ in the left side of the equation we get
\[\begin{align}
  & {{\left( x+2 \right)}^{2}} \\
 & ={{\left( -2+i\sqrt{7}+2 \right)}^{2}} \\
 & ={{\left( i\sqrt{7} \right)}^{2}} \\
 & =7\times {{i}^{2}} \\
 & =7\times \left( -1 \right) \\
 & =-7 \\
\end{align}\]
Therefore, the value $x=-2+i\sqrt{7}$ satisfies the equation ${{\left( x+2 \right)}^{2}}=-7$ as $f\left( -2+i\sqrt{7} \right)=0$.
Same thing can be said for $x=-2-i\sqrt{7}$.