Question

How do you find the zeros, real and imaginary, of `y=8x^2-4x-11` using the quadratic formula?

Answer 1

`x = (1+-sqrt(23))/4`

Explanation:

For `ax^2+bx+c=0`:

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

Therefore `a=8, b=-4, c=-11` and:

`x = (4+-sqrt((-4)^2-4*8*-11))/(2*8)`
`x = (4+-sqrt(16+352))/16`
`x = (4+-sqrt(368))/16`
`x = (4+-4sqrt(23))/16`
`x = (1+-sqrt(23))/4`