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

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Answer 1 · 2017-08-08T02:21:03

$x==1+sqrt3i$
$x=-1-sqrt3i$

Identify the values of $a,b & c$ and substitute in using the quadratic formula: $x = (-b \pm sqrt(b^2-4ac)) / (2a)$

$a=-1$

$b=-2$

$c=-4$

$x = (-(-2) \pm sqrt((-2)^2-4(-1)(-4))) / (2(-1))$

$x = (2 \pm sqrt(4-16)) / (-2)$

$x = (2 \pm sqrt(-12)) / (-2)$

$x = (2 \pm 2 sqrt(3)i) / (-2)$

$x = -1 \pm sqrt(3)i$

$x=1+sqrt3i$ , $x=-1-sqrt3ilarr$ Final solutions

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