How do you find the zeros, real and imaginary, of $y= -x^2-55x-8$ using the quadratic formula?

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Answer 1 · 2017-07-09T02:58:51

See a solution process below:

For $ax^2 + bx + c = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

Substituting $-1$ for $a$ ; $-55$ for $b$ and $-8$ for $c$ gives:

$x = (-(-55) +- sqrt((-55)^2 - (4 * -1 * -8)))/(2* -1)$

$x = (55 +- sqrt(3025 - 31))/-2$

$x = (55 +- sqrt(2994))/-2$

$x = -55/2 + sqrt(2994)/2$ and $x = -55/2 - sqrt(2994)/2$

How do you find the zeros, real and imaginary, of y= -x^2-55x-8 y= -x^2-55x-8 using the quadratic formula?