How do you find the zeros, real and imaginary, of $y = - 9 x^{2} - 28 x - 73$ $y= -9x^2-28x-73$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = - 9 x^{2} - 28 x - 73$ $y= -9x^2-28x-73$ using the quadratic formula?

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Answer 1 · 2016-04-28T07:08:11

Zeros of $y = - 9 x^{2} - 28 x - 73$ $y=-9x^2-28x-73$ are $- \frac{14}{9} - \frac{\sqrt{461}}{9} i$ $-14/9-sqrt(461)/9i$ and $- \frac{14}{9} + \frac{\sqrt{461}}{9} i$ $-14/9+sqrt(461)/9i$

Explanation:

To find zeros of $y = - 9 x^{2} - 28 x - 73$ $y=-9x^2-28x-73$ , one needs to find values of $x$ $x$ for which $y = f (x) = 0$ $y=f(x)=0$ .

For $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , solution is given by quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$ . Hence for $- 9 x^{2} - 28 x - 73 = 0$ $-9x^2-28x-73=0$ or $9 x^{2} + 28 x + 73 = 0$ $9x^2+28x+73=0$ ,

$x = \frac{- 28 \pm \sqrt{28^{2} - 4 \times 9 \times 73}}{2 \times 9}$ $x=(-28+-sqrt(28^2-4xx9xx73))/(2xx9)$

= $\frac{- 28 \pm \sqrt{784 - 2628}}{18}$ $(-28+-sqrt(784-2628))/18$

= $\frac{- 28 \pm \sqrt{- 1844}}{18}$ $(-28+-sqrt(-1844))/18$

= $\frac{- 28 \pm \sqrt{- 4 \times 461}}{18}$ $(-28+-sqrt(-4xx461))/18$

= $- \frac{14}{9} \pm \frac{\sqrt{461}}{9} i$ $-14/9+-sqrt(461)/9i$

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How do you find the zeros, real and imaginary, of $y = - 9 x^{2} - 28 x - 73$ $y= -9x^2-28x-73$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y= -9x^2-28x-73y=−9x2−28x−73y= -9x^2-28x-73 using the quadratic formula?

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Explanation:

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Impact of this question

How do you find the zeros, real and imaginary, of $y = - 9 x^{2} - 28 x - 73$ $y= -9x^2-28x-73$ using the quadratic formula?