How do you find the roots, real and imaginary, of $y=2x^2 -42x-28$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=2x^2 -42x-28$ using the quadratic formula?

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Answer 1 · 2017-07-04T17:34:45

$x = (21 +- sqrt497)/2$

Explanation:

$f(x) = 2x^2 - 42x - 28 = 0$
$f(x) = 2(x^2 - 21x - 14) = 0$
Use the improved quadratic formula in graphic form:
$D = d^2 = b^2 - 4ac = 441 + 56 = 497$ --> $d = +- sqrt497$
There are 2 real roots:
$x = -b/(2a) +- d/(2a) = 21/2 +- sqrt497/2 = (21 +- sqrt497)/2$

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How do you find the roots, real and imaginary, of $y=2x^2 -42x-28$ using the quadratic formula?

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Explanation:

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Humanities

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How do you find the roots, real and imaginary, of y=2x^2 -42x-28 y=2x^2 -42x-28 using the quadratic formula?

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Explanation:

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How do you find the roots, real and imaginary, of $y=2x^2 -42x-28$ using the quadratic formula?