Quadratic Roots C31728

1. **State the problem:** We need to find the roots of the quadratic equation $$y^2 - 3y - 18 = 0$$. 2. **Recall the quadratic formula:** For an equation $$ay^2 + by + c = 0$$, the roots are given by $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = -3$$, and $$c = -18$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-3)^2 - 4 \times 1 \times (-18) = 9 + 72 = 81$$. 5. **Find the square root of the discriminant:** $$\sqrt{\Delta} = \sqrt{81} = 9$$. 6. **Apply the quadratic formula:** $$y = \frac{-(-3) \pm 9}{2 \times 1} = \frac{3 \pm 9}{2}$$. 7. **Calculate the two roots:** - For the plus sign: $$y = \frac{3 + 9}{2} = \frac{12}{2} = 6$$. - For the minus sign: $$y = \frac{3 - 9}{2} = \frac{-6}{2} = -3$$. 8. **Final answer:** The roots are $$y = 6, -3$$.