1. **State the problem:** Solve the quadratic equation $$y + 2y + y^2 + y^2 = 3$$. 2. **Combine like terms:** $$y + 2y = 3y$$ $$y^2 + y^2 = 2y^2$$ So the equation becomes: $$2y^2 + 3y = 3$$ 3. **Rewrite the equation in standard form:** $$2y^2 + 3y - 3 = 0$$ 4. **Use the quadratic formula:** For an equation $$ay^2 + by + c = 0$$, the solutions are: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a=2$$, $$b=3$$, and $$c=-3$$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times (-3) = 9 + 24 = 33$$ 6. **Find the roots:** $$y = \frac{-3 \pm \sqrt{33}}{2 \times 2} = \frac{-3 \pm \sqrt{33}}{4}$$ 7. **Final answer:** $$y = \frac{-3 + \sqrt{33}}{4} \quad \text{or} \quad y = \frac{-3 - \sqrt{33}}{4}$$ These are the two solutions to the quadratic equation.