How do you solve \[{{\left( 2s-1 \right)}^{2}}=225\]?

Hint: To solve the above given question \[{{\left( 2s-1 \right)}^{2}}=225\] we have to use factorization method. In mathematics, factorization is the writing of a number or another mathematical object as a product of several factors, usually simpler objects of the same kind. To solve the above we will use rules of mathematics.

Complete step by step solution:
The given equation is:
\[\Rightarrow {{\left( 2s-1 \right)}^{2}}=225\]
To solve above equation first we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ this formula to open the above bracket. Here $2s$ is $a$ and $-1$ is $b$ now putting these values in the ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we get
$\begin{align}
  & \Rightarrow {{\left( 2s \right)}^{2}}+{{\left( -1 \right)}^{2}}-2\times 2s\times 1=225 \\
 & \Rightarrow 4{{s}^{2}}+1-4s=225 \\
 & \Rightarrow 4{{s}^{2}}-4s-224 = 0\\
\end{align}$
Now divide the above equation by $4$, then we get
$\Rightarrow {{s}^{2}}-s-56 = 0$
Now we will factorize it by following the some rules which are:
First we have to choose two numbers such that the product of the coefficient of ${{s}^{2}}$ and the constant term is equal to the product of those two numbers. Here we get the two numbers are $\left( -8 \right)$ and second is $7$ and the addition of these two numbers must be equal to the coefficient of $s$.
The product is $-56$ and $-1$.
Now we can write the equation as:
$\begin{align}
  & \Rightarrow {{s}^{2}}+7s-8s-56 =0 \\
 & \Rightarrow s\left( s+7 \right)-8\left( s+7 \right) =0\\
 & \Rightarrow \left( s-8 \right)\left( s+7 \right) =0 \\
\end{align}$
Now put the each above factor equals to zero then we get
$\begin{align}
  & \Rightarrow s-8=0 \\
 & \Rightarrow s=8 \\
\end{align}$ and
$\begin{align}
  & \Rightarrow s+7=0 \\
 & \Rightarrow s=-7 \\
\end{align}$
Hence we get two values of the given equation \[{{\left( 2s-1 \right)}^{2}}=225\] are $s=8,-7$.

Note: We can also solve the equation by using another method.
The given equation is:
$\Rightarrow {{\left( 2s-1 \right)}^{2}}=225$
Apply square root on both side of the equation, then we get
$\begin{align}
  & \Rightarrow \left( 2s-1 \right)=\sqrt{225} \\
 & \Rightarrow 2s-1=\pm 15 \\
\end{align}$
Now we can write the above equation as:
$\Rightarrow 2s-1=15$ and $2s-1=-15$
Now add $1$ on both side of the equation then we get,
$\Rightarrow 2s-1+1=15+1$ and $2s-1+1=-15+1$
$\Rightarrow 2s=16$ and $2s=-14$
Now divide the above equation by $2$, then we get
$s=8$ and $s=-7$
Hence we get the same answer as we solved above of the equation \[{{\left( 2s-1 \right)}^{2}}=225\] which are $s=8,-7$.