How do you find the zeros of $y = -x^2 - 1/11x + 1/7$ using the quadratic formula?

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Answer 1 · 2016-08-03T07:39:35

Zeros are $7/22-7/2sqrt(491/847)$ and $7/22+7/2sqrt(491/847)$

Quadratic formula gives the zeros of $y=ax^2+bx+c$ as $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

Hence zeros of $y=-x^2-1/11x+1/7$ are

$x=(-(-1/11)+-sqrt((-1/11)^2-4(-1)(1/7)))/(2(1/7))$

= $((1/11)+-sqrt(1/121+4/7))/(2/7)$

= $(1/11+-sqrt((7+484)/847))xx7/2$

= $7/22+-7/2sqrt(491/847)$

Hence zeros are $7/22-7/2sqrt(491/847)$ and $7/22+7/2sqrt(491/847)$

How do you find the zeros of y = -x^2 - 1/11x + 1/7 y = -x^2 - 1/11x + 1/7 using the quadratic formula?