1. We are asked to solve the quadratic equation using the product-sum method: $$x^2 - x - 90 = 0$$ 2. The product-sum method involves finding two numbers that multiply to the constant term ($-90$) and add to the coefficient of $x$ ($-1$). 3. Let's find two numbers $a$ and $b$ such that: $$a \times b = -90$$ $$a + b = -1$$ 4. By testing factor pairs of 90, we find: $$9 \times (-10) = -90$$ $$9 + (-10) = -1$$ These satisfy both conditions. 5. Rewrite the middle term using these numbers: $$x^2 + 9x - 10x - 90 = 0$$ 6. Group terms: $$(x^2 + 9x) - (10x + 90) = 0$$ 7. Factor each group: $$x(x + 9) - 10(x + 9) = 0$$ 8. Factor out the common binomial: $$(x - 10)(x + 9) = 0$$ 9. Set each factor equal to zero and solve for $x$: $$x - 10 = 0 \Rightarrow x = 10$$ $$x + 9 = 0 \Rightarrow x = -9$$ 10. Therefore, the solutions are: $$x = 10 \text{ or } x = -9$$