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

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

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Answer 1 · 2017-07-04T17:27:38

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.
$d = +- 2isqrt7$
$x = -b/(2a) +- d/(2a) = 4/-2 +- 2isqrt7/2 = - 2 +- isqrt7$

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

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

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