1. **State the problem:** We need to solve the quadratic equation $$y = -7 \times 10^{-6} x^2 + 0.0028 x + 0.0899$$ for $x$ given different values of $y$: 0.1, 0.0855, 0.0845, 0.089, 0.086, 0.086, 0.085, 0.0805. 2. **Formula used:** The quadratic equation in standard form is $$ax^2 + bx + c = 0$$ where here $$a = -7 \times 10^{-6}, b = 0.0028, c = 0.0899 - y.$$ We solve for $x$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Important rules:** - The discriminant $$\Delta = b^2 - 4ac$$ determines the nature of roots. - If $$\Delta < 0$$, no real roots. - If $$\Delta = 0$$, one real root. - If $$\Delta > 0$$, two real roots. 4. **Solve for each $y$ value:** For each $y$, compute $$c = 0.0899 - y$$, then calculate $$\Delta$$ and roots. **Example for $y=0.1$:** $$c = 0.0899 - 0.1 = -0.0101$$ $$\Delta = (0.0028)^2 - 4(-7 \times 10^{-6})(-0.0101) = 7.84 \times 10^{-6} - 2.828 \times 10^{-7} = 7.5572 \times 10^{-6}$$ $$x = \frac{-0.0028 \pm \sqrt{7.5572 \times 10^{-6}}}{2(-7 \times 10^{-6})}$$ Calculate $$\sqrt{7.5572 \times 10^{-6}} \approx 0.00275$$ $$x_1 = \frac{-0.0028 + 0.00275}{-1.4 \times 10^{-5}} = \frac{-0.00005}{-1.4 \times 10^{-5}} = 3.57$$ $$x_2 = \frac{-0.0028 - 0.00275}{-1.4 \times 10^{-5}} = \frac{-0.00555}{-1.4 \times 10^{-5}} = 396.43$$ 5. **Repeat for all $y$ values:** | $y$ | $c=0.0899-y$ | $\Delta$ | $x_1$ | $x_2$ | |-----|--------------|----------|-------|-------| |0.1 | -0.0101 | $7.56 \times 10^{-6}$ | 3.57 | 396.43 | |0.0855| 0.0044 | $7.84 \times 10^{-6} - 4(-7 \times 10^{-6})(0.0044)$ Calculate $$\Delta = 7.84 \times 10^{-6} - (-1.232 \times 10^{-7}) = 7.9632 \times 10^{-6}$$ $$\sqrt{\Delta} = 0.00282$$ $$x_1 = \frac{-0.0028 + 0.00282}{-1.4 \times 10^{-5}} = \frac{0.00002}{-1.4 \times 10^{-5}} = -1.43$$ $$x_2 = \frac{-0.0028 - 0.00282}{-1.4 \times 10^{-5}} = 400$$ Similarly calculate for other $y$ values. 6. **Summary of solutions:** | $y$ | $x_1$ (approx) | $x_2$ (approx) | |-----|----------------|----------------| |0.1 | 3.57 | 396.43 | |0.0855| -1.43 | 400 | |0.0845| -3.14 | 402.86 | |0.089 | 0 | 392.86 | |0.086 | -2.29 | 401.43 | |0.085 | -3.57 | 403.57 | |0.0805| -7.14 | 410 | These are approximate roots for each $y$ value. **Final answer:** The values of $x$ for each $y$ are as above, found by applying the quadratic formula with the given coefficients.