How do you find the zeros, real and imaginary, of y=3x^2-17x-9 using the quadratic formula?

${x}_{1} = \frac{17 - \sqrt{397}}{6}$ and ${x}_{2} = \frac{17 + \sqrt{397}}{6}$
You first need to calculate ${b}^{2} - 4 a c = \Delta$. Here, $\Delta = 289 + 4 \cdot 3 \cdot 9 = 289 + 108 = 397 > 0$ so it has 2 real roots.
The quadratic formula tells us that the roots are given by $\frac{- b \pm \sqrt{\Delta}}{2 a}$.
${x}_{1} = \frac{17 - \sqrt{397}}{6}$ and ${x}_{2} = \frac{17 + \sqrt{397}}{6}$