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

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Answer 1 · 2017-02-11T07:26:44

Zeros are $1/9-isqrt26/9$ and $1/9+isqrt26/9$ , two complex conjugate numbers

Zeros of a quadratic function $y=f(x)=-9x^2-2x-3$ are those values of $x$ for which $y=0$ .

Using quadratic formula for a function $f(x)=ax^2+bx+c$ ,

zeros are $(-b+-sqrt(b^2-4ac))/(2a)$

and hence for $y=f(x)=-9x^2-2x-3$ ,

zeros are $(-(-2)+-sqrt((-2)^2-4xx(-9)xx(-3)))/(2xx(-9))$

i.e. $(2+-sqrt(4-108))/(-18)=(2+-sqrt(-104))/18$

= $(2+-sqrt(2^2xx26xxi^2))/18$ - as $i^2=-1$

= $(2+-i2sqrt26)/18=1/9+-isqrt26/9$

Hence zeros are $1/9-isqrt26/9$ and $1/9+isqrt26/9$ , two complex conjugate numbers

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