1. **State the problem:** Solve the equation $3y^2 - 20 = 160 - 2y^2$ for $y$. 2. **Write down the formula and rules:** This is a quadratic equation. To solve, first collect like terms on one side to get a standard quadratic form $ay^2 + by + c = 0$. 3. **Step-by-step solution:** - Start with the given equation: $$3y^2 - 20 = 160 - 2y^2$$ - Add $2y^2$ to both sides to bring all $y^2$ terms to the left: $$3y^2 + 2y^2 - 20 = 160$$ - Simplify: $$5y^2 - 20 = 160$$ - Add 20 to both sides: $$5y^2 = 180$$ - Divide both sides by 5: $$y^2 = 36$$ - Take the square root of both sides: $$y = \pm 6$$ 4. **Final answer:** $$y = 6 \text{ or } y = -6$$ This means the solutions to the equation are $y=6$ and $y=-6$.