Problem of the Week

Updated at Feb 23, 2026 2:41 PM

For this week we've brought you this equation problem.

How would you solve \(\frac{{(n+2)}^{2}-3}{3}=2\)?

Here are the steps:



\[\frac{{(n+2)}^{2}-3}{3}=2\]

1
Simplify  \(\frac{{(n+2)}^{2}-3}{3}\)  to  \(-1+\frac{{(n+2)}^{2}}{3}\).
\[-1+\frac{{(n+2)}^{2}}{3}=2\]

2
Regroup terms.
\[\frac{{(n+2)}^{2}}{3}-1=2\]

3
Add \(1\) to both sides.
\[\frac{{(n+2)}^{2}}{3}=2+1\]

4
Simplify  \(2+1\)  to  \(3\).
\[\frac{{(n+2)}^{2}}{3}=3\]

5
Multiply both sides by \(3\).
\[{(n+2)}^{2}=3\times 3\]

6
Simplify  \(3\times 3\)  to  \(9\).
\[{(n+2)}^{2}=9\]

7
Take the square root of both sides.
\[n+2=\pm \sqrt{9}\]

8
Since \(3\times 3=9\), the square root of \(9\) is \(3\).
\[n+2=\pm 3\]

9
Break down the problem into these 2 equations.
\[n+2=3\]
\[n+2=-3\]

10
Solve the 1st equation: \(n+2=3\).
\[n=1\]

11
Solve the 2nd equation: \(n+2=-3\).
\[n=-5\]

12
Collect all solutions.
\[n=1,-5\]

Done
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