Question

How do you find the zeros, real and imaginary, of `y= -2x^2-12x-19` using the quadratic formula?

Answer 1

2 complex roots
`x = - 3 +- isqrt2/2`

Explanation:

`y = - 2x^2 - 12x - 19 = 0`
Use the new quadratic formula in graphic form (Socratic Search).
`D = d^2 = b^2 - 4ac = 144 - 152 = - 8`
Since D < 0, there are no real roots. There are 2 complex roots.
`d = +- 2isqrt2`
`x = -b/(2a) +- d/(2a) = 12/-4 +- (2isqrt2)/4 = - 3 +- (isqrt2)/2`