1. The problem is to solve the equation $$(-1 - m)^2 = 12 - m^2 + 1^2$$ for $m$. 2. Start by expanding $(-1 - m)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$: $$(-1 - m)^2 = (-1)^2 + 2\cdot(-1)\cdot(-m) + (-m)^2 = 1 + 2m + m^2$$ 3. Substitute the expansions back into the original equation: $$1 + 2m + m^2 = 12 - m^2 + 1$$ 4. Simplify the right side: $$12 - m^2 + 1 = 13 - m^2$$ 5. Now the equation is: $$1 + 2m + m^2 = 13 - m^2$$ 6. Bring all terms to one side to set equation to zero: $$1 + 2m + m^2 - 13 + m^2 = 0 \Rightarrow 2m + 2m^2 - 12 = 0$$ 7. Simplify: $$2m^2 + 2m - 12 = 0$$ 8. Divide the entire equation by 2: $$m^2 + m - 6 = 0$$ 9. Factor the quadratic equation: $$(m + 3)(m - 2) = 0$$ 10. Set each factor equal to zero: $$m + 3 = 0 \Rightarrow m = -3$$ $$m - 2 = 0 \Rightarrow m = 2$$ 11. Therefore, the solutions for $m$ are $$m = -3$$ or $$m = 2$$. Final answer: $$\boxed{m = -3 \text{ or } m = 2}$$