How do you find the zeros, real and imaginary, of $y=x^2+3x-9$ using the quadratic formula?

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Answer 1 · 2018-08-03T06:58:15

The zeros are $x = (-3 + 3sqrt5)/2$ and $x = (-3 - 3sqrt5)/2$ .

$y = x^2 + 3x - 9$

The quadratic formula is $x = (-b +- sqrt(b^2 - 4ac))/(2a)$ , where from our equation we know that $a = 1, b = 3, c = -9$ .

Plug them into the formula:
$x = (-3 +- sqrt(3^2 - 4(1)(-9)))/(2(1))$

$x = (-3 +- sqrt(9 + 36))/2$

$x = (-3 +- sqrt(45))/2$

$x = (-3 +- 3sqrt5)/2$

Hope this helps!

How do you find the zeros, real and imaginary, of y=x^2+3x-9y=x^2+3x-9 using the quadratic formula?