How do you solve $ 7{\left( {x + 1} \right)^2} = 161 $ ?

Hint: In order to determine the solution of the given equation, simplify the equation by dividing both sides of the equation by $ 7 $ . Take the square root on both sides of the equation and remember to place $ \pm $ for the right-hand side term as every positive number has two square roots , one is positive and other is negative. Solve the equation for $ x $ to obtain the required solution.

Complete step-by-step answer:
We are given an equation having variable $ x $ i.e. $ 7{\left( {x + 1} \right)^2} = 161 $ .
In order to solve this equation, we will first simplify the equation by dividing both side of the equation with 7
 $
  \dfrac{1}{7} \times 7{\left( {x + 1} \right)^2} = \dfrac{1}{7} \times 161 \\
  {\left( {x + 1} \right)^2} = 23 \;
  $
Taking square root on both the sides. Remember one thing, very positive value always has two roots ,one positive and negative. Place the $ \pm $ on the right -hand side of the equation while applying square root
 $ \sqrt {{{\left( {x + 1} \right)}^2}} = \pm \sqrt {23} $
As we know $ 23 $ , is not a perfect square. Even it is a prime number .
 $ x + 1 = \pm \sqrt {23} $ --------(1)
Now consider the above equation as negative and find the value of $ x $
 $
\Rightarrow x + 1 = - \sqrt {23} \\
\Rightarrow x = - 1 - \sqrt {23} \;
  $
Therefore, the solution of the given equation is $ x = - 1 + \sqrt {23} , - 1 - \sqrt {23} $ or simply we can say $ x = - 1 \pm \sqrt {23} $
So, the correct answer is “ $ x = - 1 \pm \sqrt {23} $ ”.

Note: 1. Don’t forget to cross-check your answer at the end.
2.Since every positive number has two square roots, one is positive and another is negative. So place the $ \pm $ for every positive square root value.
3.The equation given was quadratic in nature, so we obtained two solutions for $ x $ as every quadratic equation has two roots.