How do you find the zeros, real and imaginary, of $y=- 5x^2-2x+10$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=- 5x^2-2x+10$ using the quadratic formula?

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Answer 1 · 2016-07-21T04:42:35

$x = (-1 +- sqrt51)/5$

Explanation:

$y = - 5x^2 - 2x + 10 = 0$
Use the new improved quadratic formula (Socratic Search)
$D = d^2 = b^2 - 4ac = 4 + 200 = 204 = 4(51)$ --> $d = +- 2sqrt51$
There are 2 real roots:
$x = -b/(2a) +- d/(2a) = 2/-10 +- (2sqrt51)/-10 = -1/5 +- sqrt51/5 =$
$x = (-1 +- sqrt51)/5$

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How do you find the zeros, real and imaginary, of $y=- 5x^2-2x+10$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of y=- 5x^2-2x+10 y=- 5x^2-2x+10 using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=- 5x^2-2x+10$ using the quadratic formula?