How do you find the zeros, real and imaginary, of $y= -x^2-6x-4$ using the quadratic formula?

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Answer 1 · 2016-01-14T20:56:26

$x=-3+-sqrt5$

For any quadratic equation $y=ax^2+bx+c$ , the zeroes are found through the formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

For the given quadratic equation:

$a=-1$
$b=-6$
$c=-4$

Plug these into the quadratic formula.

$x=(-(-6)+-sqrt((-6)^2-(4xx-1xx-4)))/(2xx-1)$

Which simplifies to be

$x=(6+-sqrt(36-16))/(-2)$

$x=(-6+-sqrt20)/2$

$x=(-6+-2sqrt5)/2$

$x=-3+-sqrt5$

How do you find the zeros, real and imaginary, of y= -x^2-6x-4y= -x^2-6x-4 using the quadratic formula?