How do you find the zeros, real and imaginary, of $y=x^2-x-9$ using the quadratic formula?

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Answer 1 · 2016-01-23T00:17:44

$x=(1+-sqrt37)/2$

The quadratic formula find the zeroes of a quadratic equation in the form $y=ax^2+bx+c$ through

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

In the given quadratic equation, we know that $a=1,b=-1,c=-9$ . Plugging these into the quadratic formula gives

$x=(-(-1)+-sqrt((-1)^2-(4xx-9xx1)))/(2xx1)$

$x=(1+-sqrt(1-(-36)))/2$

$x=(1+-sqrt37)/2$

Thus, the two times when the function equal $0$ are at $x=(1+sqrt37)/2$ and $x=(1-sqrt37)/2$ .

How do you find the zeros, real and imaginary, of y=x^2-x-9y=x^2-x-9 using the quadratic formula?