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

1 Answer
Jul 8, 2017

Answer:

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

Explanation:

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.