1. **State the problem:** Solve the quadratic equation derived from the expression $(2 - y)^2 = -3y(2 - y)$. 2. **Rewrite the equation:** Expand both sides: $$ (2 - y)^2 = -3y(2 - y) $$ $$ 4 - 4y + y^2 = -6y + 3y^2 $$ 3. **Bring all terms to one side:** $$ 4 - 4y + y^2 + 6y - 3y^2 = 0 $$ $$ 4 + 2y - 2y^2 = 0 $$ 4. **Simplify the equation:** $$ -2y^2 + 2y + 4 = 0 $$ 5. **Divide entire equation by -2 to simplify:** $$ \cancel{-2}y^2 + \cancel{-2}y + \cancel{-2}2 = 0 \Rightarrow y^2 - y - 2 = 0 $$ 6. **Factor the quadratic:** $$ y^2 - y - 2 = (y - 2)(y + 1) = 0 $$ 7. **Solve for $y$:** $$ y - 2 = 0 \Rightarrow y = 2 $$ $$ y + 1 = 0 \Rightarrow y = -1 $$ **Final answer:** The solutions are $y = 2$ and $y = -1$.