How do you find the zeros, real and imaginary, of $y= -2x^2-x-19$ using the quadratic formula?

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Answer 1 · 2016-05-29T07:39:14

Zeros of $f(x)$ are ${-1/4+isqrt151/4,-1/4-isqrt151/4}$

To find zeros of $f(x)=-2x^2-x-19$ , we need roots of the equation $-2x^2-x-19=0$ , which can be obtained using quadratic formula.

As the roots of $ax^2+bx+c=0$ are $x=(-b+-sqrt(b^2-4ac))/(2a)$

Hence, roots of zeros of $-2x^2-x-19=0$ are

$x=(-(-1)+-sqrt((-1)^2-4(-2)(-19)))/(2(-2))$

= $(1+-sqrt(1-152))/(-4)=-1/4+-isqrt151/4$

Zeros of $f(x)$ are ${-1/4+isqrt151/4,-1/4-isqrt151/4}$

How do you find the zeros, real and imaginary, of y= -2x^2-x-19 y= -2x^2-x-19 using the quadratic formula?