How do you solve ${{\left( x-6 \right)}^{2}}=25$ ?

Hint: To solve the above equation, we are going to first subtract 25 on both the sides. Then we will require the following algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. After applying this algebraic identity we can easily find the values of x.

Complete step-by-step solution:
The quadratic equation given in the above problem which we have to solve is as follows:
${{\left( x-6 \right)}^{2}}=25$
Subtracting 25 on both the sides of the above equation we get,
$\begin{align}
  & {{\left( x-6 \right)}^{2}}-25=25-25 \\
 & \Rightarrow {{\left( x-6 \right)}^{2}}-25=0 \\
\end{align}$
In the above equation, we can write 25 as ${{5}^{2}}$ so we are writing 25 in this manner we get,
${{\left( x-6 \right)}^{2}}-{{5}^{2}}=0$
As you can see that the expression written in the L.H.S of the above equation is of the form ${{a}^{2}}-{{b}^{2}}$ so we can use the identity of ${{a}^{2}}-{{b}^{2}}$ which is equal to:
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
$\begin{align}
  & \Rightarrow {{\left( x-6 \right)}^{2}}-{{5}^{2}}=0 \\
 & \Rightarrow \left( x-6-5 \right)\left( x-6+5 \right)=0 \\
 & \Rightarrow \left( x-11 \right)\left( x-1 \right)=0 \\
\end{align}$
To find the value of x in the above equation, we are going to equate each bracket to 0 and we get,
$\begin{align}
  & x-11=0 \\
 & \Rightarrow x=11; \\
\end{align}$
$\begin{align}
  & x-1=0 \\
 & \Rightarrow x=1 \\
\end{align}$
From the above solutions, we have got the solutions of x as 1 and 11.
Hence, we have solved the equation given in the above problem and its solutions are 1 and 11.

Note: To check the solutions of the quadratic equation, we are going to substitute the values of x which we have obtained above in the given equation and then check whether these values of x are either satisfying the given equation or not.
The values of x which we have obtained above are:
1 and 11
Substituting the value of x as 1 in the above equation we get,
$\begin{align}
  & {{\left( x-6 \right)}^{2}}=25 \\
 & \Rightarrow {{\left( 1-6 \right)}^{2}}=25 \\
 & \Rightarrow {{\left( -5 \right)}^{2}}=25 \\
 & \Rightarrow 25=25 \\
\end{align}$
From the above, it is proven that the value of x equal to 1 is satisfying the given equation.
Similarly, you can check the other value of x (which is equal to 11) whether this value is satisfying the given equation or not.