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

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Answer 1 · 2016-03-18T03:22:49

$x = -3 +- sqrt11$

$D = d^2 = b^2 - 4ac = 36 + 8 = 44$ --> $d = +- 2sqrt11$
There are 2 real roots:
$x = 6/-2 +- 2sqrt11/-2 = - 3 +- sqrt11$

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