How do you find the zeros, real and imaginary, of $y=-x^2-2x+33$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=-x^2-2x+33$ using the quadratic formula?

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Answer 1 · 2016-06-25T03:51:04

Two real roots:
$- 1 +- sqrt34$

Explanation:

Use the improved quadratic formula (Socratic Search)
$y= -x^2 - 2x + 33 = 0$
$D = d^2 = b^2 - 4ac = 4 + 132 = 136$ --> $d = +- 2sqrt34$
There are 2 real roots:
$x = -b/(2a) +- d/(2a) = 2/-2 +- (2sqrt34)/-2 = -1 +- sqrt34$

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How do you find the zeros, real and imaginary, of $y=-x^2-2x+33$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of y=-x^2-2x+33y=-x^2-2x+33 using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=-x^2-2x+33$ using the quadratic formula?