1. **State the problem:** Find all values of $m$ for which the equation $$4 = -(m - 1)w^2 - 8w$$ has exactly one real solution in $w$. 2. **Rewrite the equation:** Move all terms to one side to get a quadratic in $w$: $$-(m - 1)w^2 - 8w - 4 = 0$$ 3. **Identify coefficients:** The quadratic is of the form $aw^2 + bw + c = 0$ where: $$a = -(m - 1) = 1 - m, \quad b = -8, \quad c = -4$$ 4. **Condition for one real solution:** A quadratic has exactly one real solution if its discriminant $\Delta$ is zero: $$\Delta = b^2 - 4ac = 0$$ 5. **Calculate the discriminant:** $$b^2 - 4ac = (-8)^2 - 4(1 - m)(-4) = 64 + 16(1 - m) = 64 + 16 - 16m = 80 - 16m$$ 6. **Set discriminant to zero and solve for $m$:** $$80 - 16m = 0$$ $$16m = 80$$ $$m = \frac{80}{16} = 5$$ 7. **Check the leading coefficient:** For the equation to be quadratic, $a = 1 - m \neq 0$. At $m=5$, $a = 1 - 5 = -4 \neq 0$, so the equation is quadratic. **Final answer:** $$m = 5$$