If $3{{\left( x-2 \right)}^{2}}=507$, then ‘x’ can be_______

Hint: To find the value of ‘x’ we had to solve the above equation. So we would try to reduce it to a simple or general look of equations which we can solve through general identities.

Complete step by step answer:
Moving ahead with the question,
As In the given equation$3{{\left( x-2 \right)}^{2}}=507$, we had${{\left( x-2 \right)}^{2}}$, which looks like we can have some square type term in equation, of which we know the identity, so let us try to have it separately, which we can do by taking the LHS side numerical ‘3’ to RHS side. So we will get;
$\begin{align}
  & 3{{\left( x-2 \right)}^{2}}=507 \\
 & {{\left( x-2 \right)}^{2}}=\dfrac{507}{3} \\
\end{align}$
On simplifying it further;
${{\left( x-2 \right)}^{2}}=169$
As we know that 169 is a square of 13, so let us replace it, so that both sides we can have the exponential value, which will make us easy to find out value of ‘x’. so we will get;
${{\left( x-2 \right)}^{2}}={{\left( 13 \right)}^{2}}$
As both sides are square then we can take square of 13 to LHS side, which will make it type of\[{{a}^{2}}-{{b}^{2}}\]which we can further write it as$\left( a-b \right)\left( a+b \right)$, so by comparing we can say that ‘a’ is equal to$x-2$and ‘y’ is equal to$13$. So by putting the value we will get;
$\begin{align}
  & {{\left( x-2 \right)}^{2}}={{\left( 13 \right)}^{2}} \\
 & {{\left( x-2 \right)}^{2}}-{{\left( 13 \right)}^{2}}=0 \\
 & \left( x-2-13 \right)\left( x-2+13 \right)=0 \\
 & \left( x-15 \right)\left( x+11 \right)=0 \\
\end{align}$
By comparing it we can say that any of two factor i.e. $\left( x-15 \right)$or$\left( x+11 \right)$must be zero. So by it we can say that;
$\left( x-15 \right)=0$ or $\left( x+11 \right)=0$
On solving we will get;
$x=15$ or $x=-11$
So from here we got two values of ‘x’ i.e.$x=15$and$x=-11$.
Hence the answer is$15,-11$.

Note: So solve any equation there are many ways to simplify it, and method will give the same answer, it depends on us which method suits us. So if someone gets the answer to this equation from another method does not mean this method is wrong.