1. **State the problem:** Solve the equation $$y^2 + 2y - 21 = 0$$ for $y$. 2. **Formula and rules:** This is a quadratic equation of the form $$ay^2 + by + c = 0$$ where $a=1$, $b=2$, and $c=-21$. We use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The discriminant $$\Delta = b^2 - 4ac$$ determines the nature of the roots. 3. **Calculate the discriminant:** $$\Delta = 2^2 - 4 \times 1 \times (-21) = 4 + 84 = 88$$ Since $$\Delta > 0$$, there are two distinct real roots. 4. **Apply the quadratic formula:** $$y = \frac{-2 \pm \sqrt{88}}{2} = \frac{-2 \pm 2\sqrt{22}}{2} = -1 \pm \sqrt{22}$$ 5. **Final answer:** $$y = -1 + \sqrt{22} \quad \text{or} \quad y = -1 - \sqrt{22}$$ These are the two solutions to the equation.