# How do you find the zeros of  y = 7x^2 + x -2  using the quadratic formula?

The roots are $\frac{- 1 \pm \sqrt{57}}{14}$.
You first calculate $\Delta = {b}^{2} - 4 a c$, which here equals to $1 - 4 \cdot 7 \cdot \left(- 2\right) = 57$. It's positive so $y$ has two roots.
The quadratic formula says that the roots of a trinomial are (-b +- sqrt(Delta))/(2a. So the two roots of $y$ are $\frac{- 1 - \sqrt{57}}{14}$ and $\frac{- 1 + \sqrt{57}}{14}$.