1. **State the problem:** We have an isosceles triangle ABC with AB = AC, and angles ABC and ACB given by expressions in terms of $x$: $\angle ABC = 3x^2 - 2x + 4$ degrees and $\angle ACB = 9x - 6$ degrees. We need to find the two possible sets of the three angles of triangle ABC. 2. **Recall properties and formulas:** In an isosceles triangle with AB = AC, the base angles are equal, so $\angle ABC = \angle ACB$. 3. **Set up the equation:** Since $\angle ABC = \angle ACB$, we have: $$3x^2 - 2x + 4 = 9x - 6$$ 4. **Solve the quadratic equation:** Move all terms to one side: $$3x^2 - 2x + 4 - 9x + 6 = 0$$ $$3x^2 - 11x + 10 = 0$$ 5. **Factor or use quadratic formula:** The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=-11$, $c=10$. Calculate the discriminant: $$\Delta = (-11)^2 - 4 \times 3 \times 10 = 121 - 120 = 1$$ Calculate roots: $$x = \frac{11 \pm \sqrt{1}}{6}$$ $$x_1 = \frac{11 + 1}{6} = 2$$ $$x_2 = \frac{11 - 1}{6} = \frac{10}{6} = \frac{5}{3} \approx 1.6667$$ 6. **Find the angles for each $x$:** For $x=2$: $$\angle ABC = 3(2)^2 - 2(2) + 4 = 3 \times 4 - 4 + 4 = 12 - 4 + 4 = 12$$ $$\angle ACB = 9(2) - 6 = 18 - 6 = 12$$ Since $\angle ABC = \angle ACB = 12^\circ$, the third angle $\angle BAC$ is: $$180 - 12 - 12 = 156^\circ$$ For $x=\frac{5}{3}$: $$\angle ABC = 3\left(\frac{5}{3}\right)^2 - 2\left(\frac{5}{3}\right) + 4 = 3 \times \frac{25}{9} - \frac{10}{3} + 4 = \frac{75}{9} - \frac{10}{3} + 4 = \frac{25}{3} - \frac{10}{3} + 4 = \frac{15}{3} + 4 = 5 + 4 = 9$$ $$\angle ACB = 9 \times \frac{5}{3} - 6 = 15 - 6 = 9$$ The third angle $\angle BAC$ is: $$180 - 9 - 9 = 162^\circ$$ 7. **Final answer:** The two possible sets of angles for triangle ABC are: - $\{12^\circ, 12^\circ, 156^\circ\}$ - $\{9^\circ, 9^\circ, 162^\circ\}$ These satisfy the properties of an isosceles triangle and the given angle expressions.