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#