Question

How do you find the zeros, real and imaginary, of `y= -x^2-6x+2` using the quadratic formula?

Answer 1

`x = -3 +- sqrt11`

Explanation:

`D = d^2 = b^2 - 4ac = 36 + 8 = 44` --> `d = +- 2sqrt11`
There are 2 real roots:
`x = 6/-2 +- 2sqrt11/-2 = - 3 +- sqrt11`