How do you solve ${{x}^{2}}+28x+196=0$?

Hint: We use the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ to form a square for the left side of the equation ${{x}^{2}}+28x+196=0$. Then we take the square root on both sides of the equation. From that we subtract 14 to the both sides to find the value of $x$ for ${{x}^{2}}+28x+196=0$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}+28x+196=0$.
We use the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ to form the square.
$\begin{align}
  & {{x}^{2}}+28x+196=0 \\
 & \Rightarrow {{x}^{2}}+2\times 14\times x+{{14}^{2}}=0 \\
 & \Rightarrow {{\left( x+14 \right)}^{2}}=0 \\
\end{align}$
We interchanged the numbers for $a=x,b=14$.
Now we have a quadratic equation ${{\left( x+14 \right)}^{2}}=0$.
We need to find the solution of the given equation ${{\left( x+14 \right)}^{2}}=0$.
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.
$\begin{align}
  & \sqrt{{{\left( x+14 \right)}^{2}}}=0 \\
 & \Rightarrow \left( x+14 \right)=0 \\
 & \Rightarrow x=-14 \\
\end{align}$

The given quadratic equation has two solutions and they are $x=-14$.

Note: We can also break the given equation in the form of a grouping method where we break the term $28x$ in two parts of $14x$ and $14x$. We take common terms and form multiplication.
$\begin{align}
  & {{x}^{2}}+28x+196 \\
 & \Rightarrow {{x}^{2}}+14x+14x+196 \\
 & \Rightarrow x\left( x+14 \right)+14\left( x+14 \right) \\
 & \Rightarrow \left( x+14 \right)\left( x+14 \right) \\
 & \Rightarrow {{\left( x+14 \right)}^{2}} \\
\end{align}$
The solution after that follows the main method and we get $x=-14$.