How do you find the zeros, real and imaginary, of $y=-x^2 -2x +5$ using the quadratic formula?

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Answer 1 · 2017-11-05T11:27:20

The quadratic formula says that the zeros of a quadratic function can be calculated as:

$x_1=(-b-sqrt(Delta))/(2a)$

and

where

Here we have:

$Delta=(-2)^2-4*(-1)*5=4+20=24$

$sqrt(Delta)=sqrt(24)=2sqrt(6)$

The discriminant $(Delta)$ is greater than zero, so the function has 2 different real zeros:

$x_1=(2-2sqrt(6))/(-2)=-1+sqrt(6)$

$x_1=(2+2sqrt(6))/(-2)=-1-sqrt(6)$