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

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

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Answer 1 · 2016-07-30T15:45:55

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.
$d = +- 2isqrt2$
$x = -b/(2a) +- d/(2a) = 12/-4 +- (2isqrt2)/4 = - 3 +- (isqrt2)/2$

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

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

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