Solve each equation. Check your solutions.

${\left(x-5\right)}^{\mathbf{2}}\mathbf{=}\mathbf{121}$

A quadratic equation is a second order equation written as$a{x}^2+bx+c=0$ where a, b, and c are coefficients of real numbers and $a\ne 0$.

A quadratic equation generally has two roots, which can be equal to each other. Some quadratic expressions can be factored, which means that the equation can be written as:

$a\left(x-x_1\right)\left(x-x_2\right)=0$

where$x_1$ and$x_2$ are the two roots of the equation.

Factoring an algebraic expression means writing the expression as a product of factors.

To verify whether the factors are correct or not, multiply them and check if the result is the original algebraic expression.

Algebraic expressions can be factorized using the common factor method, regrouping like terms together, and also by using algebraic identities.

Factorize the given polynomial by using the formula: $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

$\begin{array}{c}{\left(x-5\right)}^2=121\\ {\left(x-5\right)}^2-121=0\\ {\left(x-5\right)}^2-{\left(11\right)}^2=0\\ \left(x-5-11\right)\left(x-5+11\right)=0\\ \left(x-16\right)\left(x+6\right)=0\\ x-16=0\text{or}x+6=0\\ x=16\text{or}x=-6\\ x=-6,16\end{array}$

Therefore, the solution to the given equation is $x=-6,16$.