Question

How do you find the zeroes for `y=9x^2+6x-1`?

Answer 1

-0.805 or 0.138

Explanation:

Assuming `y=0` you can use the quadratic formula

`x=(-b±sqrt(b^2-4ac))/(2a)`

to find the roots of a quadratic.
A quadratic is defined as `ax^2+bx+c` so for your case you may plug in the respective values of `a`, `b`, and `c`:

`a=9`
`b=6`
`c=-1`

Then the quadratic formula would look like this:

`x=(-6±sqrt(6^2-4*9*-1))/(2*9)`

Due to the `±` sign we then have two variants for 2 answers (2 is also the degree of the polynomial which determines the amount of roots/answers of the polynomial)

`x=(-6+sqrt(6^2-4*9*-1))/(2*9)=0.138071187458`
`x=(-6-sqrt(6^2-4*9*-1))/(2*9)=-0.804737854124`

I rounded the answers to `x=-0.805, x=0.138`
To check your answer a graph can be utilized

graph{9x^2+6x-1 [-5.285, 4.58, -2.74, 2.193]}

:-)