How do you find the zeros, real and imaginary, of $y=-x^2 -2x +8$ using the quadratic formula?

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Answer 1 · 2016-02-14T11:03:27

$x_1 = -4$
$x_2 = 2$

Given a quadratic equation:
$y = ax^2 + bx + c$

We can find its zeros using the Quadratic Formula:
$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

$y = -x^2 - 2x +8$

$=> x = (-(-2) +- sqrt((-2)^2 - 4*(-1)*(8)))/(2(-1))$

$=> x = (2 +- sqrt(4 - (-32)))/-2$

$=> x = (2 +- sqrt 36)/-2$

$=> x = (2 +- 6)/-2$

$=> x_1 = 8/-2 = -4$

$=> x_2 = -4/-2 = 2$

How do you find the zeros, real and imaginary, of y=-x^2 -2x +8y=-x^2 -2x +8 using the quadratic formula?