Question

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

Answer 1

2 imaginary roots:
`x = - 2 +- isqrt7`

Explanation:

`y = - x^2 - 4x - 11 = 0`
Use the improved quadratic formula in graphic form (Google Search)
`D = d^2 = b^2 - 4ac = 16 - 44 = - 28`
Since D < 0, there are 2 imaginary roots.
`d = +- 2isqrt7`
`x = -b/(2a) +- d/(2a) = 4/-2 +- 2isqrt7/2 = - 2 +- isqrt7`