1. **State the problem:** Factor the quadratic equation $$-x^2 + 3x + 10 = 0$$. 2. **Rewrite the equation:** It is easier to factor if the leading coefficient is positive. Multiply the entire equation by $$-1$$: $$x^2 - 3x - 10 = 0$$ 3. **Identify coefficients:** Here, $$a = 1$$, $$b = -3$$, and $$c = -10$$. 4. **Find two numbers that multiply to $$a \times c = 1 \times (-10) = -10$$ and add to $$b = -3$$. These numbers are $$2$$ and $$-5$$ because $$2 \times (-5) = -10$$ and $$2 + (-5) = -3$$. 5. **Rewrite the middle term using these numbers:** $$x^2 + 2x - 5x - 10 = 0$$ 6. **Group terms:** $$(x^2 + 2x) + (-5x - 10) = 0$$ 7. **Factor each group:** $$x(x + 2) - 5(x + 2) = 0$$ 8. **Factor out the common binomial:** $$(x - 5)(x + 2) = 0$$ 9. **Recall the original equation was multiplied by $$-1$$, so the factors of the original equation are:** $$-(x - 5)(x + 2) = 0$$ 10. **Final answer:** The factored form of $$-x^2 + 3x + 10 = 0$$ is $$-(x - 5)(x + 2) = 0$$.