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.

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