## ihatealgebrasomuch 3 years ago What would be the best way to solve 3d^2+6d-1=0

Use the quadratic formula as it works with every quadratic (regardless of solution type) $\Large d = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $\Large d = \frac{-(6)\pm\sqrt{(6)^2-4(3)(-1)}}{2(3)}$ $\Large d = \frac{-6\pm\sqrt{36-(-12)}}{6}$ $\Large d = \frac{-6\pm\sqrt{48}}{6}$ $\Large d = \frac{-6+\sqrt{48}}{6} \ \text{or} \ d = \frac{-6-\sqrt{48}}{6}$ $\Large d = \frac{-6+4\sqrt{3}}{6} \ \text{or} \ d = \frac{-6-4\sqrt{3}}{6}$ $\Large d = \frac{-3+2\sqrt{3}}{3} \ \text{or} \ d = \frac{-3-2\sqrt{3}}{3}$ So the solutions are $\Large d = \frac{-3+2\sqrt{3}}{3} \ \text{or} \ d = \frac{-3-2\sqrt{3}}{3}$