How do you find the zeros, real and imaginary, of $y=5(x-3)^2-56$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=5(x-3)^2-56$ using the quadratic formula?

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Answer 1 · 2016-01-10T23:12:38

$x = 3 +- 3,35$

Explanation:

Develop $y = 5(x^2 - 6x + 9) - 56 = 5x^2 - 30x + 45 - 56$
$y = 5x^2 - 30x - 11 = 0$
$D = d^2 = b^2 - 4ac = 900 + 220 = 1120$ --> $d = +- 33.47$ .
There are 2 real roots:
$x = 30/10 +- 33.47/10 = 3 +- 3.35$
x1 = 6.35 and x2 = 0.35

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How do you find the zeros, real and imaginary, of $y=5(x-3)^2-56$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of y=5(x-3)^2-56 y=5(x-3)^2-56 using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=5(x-3)^2-56$ using the quadratic formula?