Question

How do you solve `3z^2 + z - 1 = 0` using the quadratic formula?

Answer 1

`x=(-1+-sqrt13)/6`

Explanation:

The Quadratic Formula is

`x=(-b+-sqrt(b^2-4ac))/(2a)`

In your equation which is in the general form
`az^2+bz+c=0`

`a=3, b=1, c=-1`

So putting it into the quadratic formula,

`x=(-1+-sqrt(1+12))/6`
`x=(-1+-sqrt13)/6`