The given equation is:

$c^2 = 121$

To find the solution(s), take the square root of both sides of the equation:

$c = \pm \sqrt{121}$

Since $\sqrt{121} = 11$, we have:

$c = \pm 11$

Therefore, the solutions are:

$c = 11 \quad \text{or} \quad c = -11$

Would you like any further details or clarifications? Here are 5 related questions for further exploration:

1. How do we solve equations with higher powers, such as $c^3 = 8$?
2. What happens if we solve $c^2 = -121$?
3. Can we solve $c^2 = 0$ the same way?
4. What is the difference between a positive and negative square root?
5. How does this approach apply to solving quadratic equations?

Tip: Always consider both the positive and negative roots when solving $x^2 = a$, unless further restrictions are given.