How do you find the zeros of $y = -x^2 - 12x + 7$ using the quadratic formula?

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Answer 1 · 2015-11-28T07:40:58

The zeros are at
$x = -6+sqrt(43)$ and $x = -6 - sqrt(43)$

The quadratic formula states that
$ax^2 + bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)$

We are looking for the values of $x$ at which

$-x^2 - 12x + 7 = 0$

Then we can apply the quadratic formula with
$a=-1, b=-12, c=7$
to obtain

$x = (-(-12)+-sqrt((-12)^2-4(-1)(7)))/(2(-1))$

$= (12+-sqrt(144 + 28))/(-2)$

$= -(12+-sqrt(172))/2$

$= -(12 +- 2sqrt(43))/2$

$= -6 +- sqrt(43)$

Thus the zeros are at
$x = -6+sqrt(43)$ and $x = -6 - sqrt(43)$

How do you find the zeros of y = -x^2 - 12x + 7 y = -x^2 - 12x + 7 using the quadratic formula?