How do you find the zeros, real and imaginary, of $y= -9x^2-28x-3$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y= -9x^2-28x-3$ using the quadratic formula?

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Answer 1 · 2018-07-26T18:06:43

The zeros are $x = -3, -1/9$ .

Explanation:

$y = -9x^2 - 28x - 3$

This equation is in standard quadratic form, or $y = color(red)(a)x^2 + color(blue)(b)x + color(magenta)(c)$ , where $color(red)(a = -9)$ , $color(blue)(b = -28)$ , and $color(magenta)(c = -3)$ .

The quadratic formula is $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - 4color(red)(a)color(magenta)(c)))/(2color(red)(a))$ .

Plug in the values and solve:
$x = (-(color(blue)(-28)) +- sqrt((color(blue)(-28))^2 - 4(color(red)(-9))(color(magenta)(-3))))/(2(color(red)(-9)))$ :

$x = (28 +- sqrt(784 - 108))/(-18)$

$x = (28 +- sqrt(676))/(-18)$

$x = (28 +- 26)/(-18)$

$x = (28+26)/(-18)$ , $(28-26)/(-18)$

$x = 54/-18$ , $2/-18$

$x = -3, -1/9$

Hope this helps!

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How do you find the zeros, real and imaginary, of $y= -9x^2-28x-3$ using the quadratic formula?

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Explanation:

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How do you find the zeros, real and imaginary, of y= -9x^2-28x-3y= -9x^2-28x-3 using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y= -9x^2-28x-3$ using the quadratic formula?