1. **State the problem:** Solve the quadratic equation $$(k+2)x^2 - 2kx + k - 1 = 0$$ for $x$. 2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = k+2$, $b = -2k$, and $c = k-1$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-2k)^2 - 4(k+2)(k-1) = 4k^2 - 4(k+2)(k-1)$$ 4. **Expand and simplify the discriminant:** $$4k^2 - 4[(k+2)(k-1)] = 4k^2 - 4(k^2 - k + 2k - 2) = 4k^2 - 4(k^2 + k - 2)$$ $$= 4k^2 - 4k^2 - 4k + 8 = -4k + 8$$ 5. **Write the solutions using the quadratic formula:** $$x = \frac{-(-2k) \pm \sqrt{-4k + 8}}{2(k+2)} = \frac{2k \pm \sqrt{8 - 4k}}{2(k+2)}$$ 6. **Simplify the square root and denominator:** $$\sqrt{8 - 4k} = \sqrt{4(2 - k)} = 2\sqrt{2 - k}$$ So, $$x = \frac{2k \pm 2\sqrt{2 - k}}{2(k+2)} = \frac{2(k \pm \sqrt{2 - k})}{2(k+2)} = \frac{k \pm \sqrt{2 - k}}{k+2}$$ 7. **Final answer:** $$\boxed{x = \frac{k \pm \sqrt{2 - k}}{k+2}}$$ Note: The solutions are real only if the discriminant is non-negative, i.e., $$2 - k \geq 0 \Rightarrow k \leq 2$$ and the denominator $$k+2 \neq 0 \Rightarrow k \neq -2$$.