How do you find the zeros, real and imaginary, of #y=-4x^2-4x-16# using the quadratic formula?

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Answer 1 · 2017-07-08T17:20:58

#-(1+-isqrt(15))/(2)#

The standard form of a quadratic equation is #y=ax^2+bx+c#. Based on this, you know that #a=-4#, #b=-4#, and #c=-16#.

The quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a)#. Substitute the values above into the formula.

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

#x=(-(-4)+-sqrt((-4)^2-4(-4)(-16)))/(2(-4))#

#x=(4+-sqrt(-240))/(-8)#

#x=(4+-4isqrt(15))/(-8)#

#x=(1+-isqrt(15))/(-2) -> -(1+-isqrt(15))/(2)#

There are no real zeros.