How do you solve \[{{\left( x+1 \right)}^{2}}-\dfrac{9}{4}=0\]?

Hint: In this problem, we have to solve and find the value of x. We can first take the constant term from the left-hand side to the right-hand side of the equation or we can add the constant term on both sides to get cancelled. We can then take square roots on both sides. We should know some square terms to solve this problem. By using the square terms, we can simplify the resulting step to find the value of x.

Complete step by step solution:
We know that the given equation to be solved is,
\[{{\left( x+1 \right)}^{2}}-\dfrac{9}{4}=0\]
We can now take the constant term from the left-hand side to the right-hand side of the equation, we get
\[\Rightarrow {{\left( x+1 \right)}^{2}}=\dfrac{9}{4}\]
We can now take square root on both sides we get,
\[\Rightarrow {{\sqrt{\left( x+1 \right)}}^{2}}=\pm \sqrt{\dfrac{9}{4}}\]
Now we can simplify the above step by cancelling the square and the square root on the left-hand side and by simplifying using root values in the right-hand side, we get
\[\begin{align}
  & \Rightarrow x+1=\pm \dfrac{3}{2} \\
 & \because \sqrt{9}=3,\sqrt{4}=2 \\
\end{align}\]
Now we can subtract the number on both the left-hand side and the right-hand side, we get
\[\Rightarrow x=-1\pm \dfrac{3}{2}\]
Now we can further simplify the above step, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-2+3}{2}=\dfrac{1}{2} \\
 & \Rightarrow x=\dfrac{-2-3}{2}=\dfrac{-5}{2} \\
\end{align}\]
Therefore, the value of \[x=\dfrac{1}{2},-\dfrac{5}{2}\]

Note: Students make mistakes while writing the root values which should be concentrated. We should also concentrate in the plus or minus symbol before the root. We know that to take square root on both sides we should know some square values