1. **State the problem:** Solve the equation $$0 = -3x^5 - 10x^3 + 45x$$ for $x$. 2. **Identify the formula and rules:** This is a polynomial equation. To solve it, we factor the polynomial and set each factor equal to zero. 3. **Factor the polynomial:** First, factor out the greatest common factor (GCF), which is $-x$: $$0 = -x(3x^4 + 10x^2 - 45)$$ 4. **Set each factor equal to zero:** - For $-x = 0$, we get: $$x = 0$$ - For $3x^4 + 10x^2 - 45 = 0$, let $y = x^2$ to reduce the quartic to a quadratic in $y$: $$3y^2 + 10y - 45 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=10$, and $c=-45$. Calculate the discriminant: $$\Delta = 10^2 - 4 \times 3 \times (-45) = 100 + 540 = 640$$ So, $$y = \frac{-10 \pm \sqrt{640}}{6} = \frac{-10 \pm 8\sqrt{10}}{6}$$ Simplify: $$y = \frac{-10 + 8\sqrt{10}}{6} \quad \text{or} \quad y = \frac{-10 - 8\sqrt{10}}{6}$$ 6. **Evaluate the roots for $y$:** - $y_1 = \frac{-10 + 8\sqrt{10}}{6}$ is approximately positive (since $8\sqrt{10} \approx 25.3$), so $y_1 > 0$. - $y_2 = \frac{-10 - 8\sqrt{10}}{6}$ is negative. Since $y = x^2$, $y$ must be non-negative. So discard $y_2$. 7. **Find $x$ from $y_1$:** $$x^2 = y_1 = \frac{-10 + 8\sqrt{10}}{6}$$ Therefore, $$x = \pm \sqrt{\frac{-10 + 8\sqrt{10}}{6}}$$ 8. **Final solutions:** $$x = 0, \quad x = \pm \sqrt{\frac{-10 + 8\sqrt{10}}{6}}$$