Question

How do you find the zeros, real and imaginary, of `y= -9x^2-28x-3` using the quadratic formula?

Answer 1

The zeros are `x = -3, -1/9`.

Explanation:

`y = -9x^2 - 28x - 3`

This equation is in standard quadratic form, or `y = color(red)(a)x^2 + color(blue)(b)x + color(magenta)(c)`, where `color(red)(a = -9)`, `color(blue)(b = -28)`, and `color(magenta)(c = -3)`.

The quadratic formula is `x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - 4color(red)(a)color(magenta)(c)))/(2color(red)(a))`.

Plug in the values and solve:
`x = (-(color(blue)(-28)) +- sqrt((color(blue)(-28))^2 - 4(color(red)(-9))(color(magenta)(-3))))/(2(color(red)(-9)))`:

`x = (28 +- sqrt(784 - 108))/(-18)`

`x = (28 +- sqrt(676))/(-18)`

`x = (28 +- 26)/(-18)`

`x = (28+26)/(-18)`, `(28-26)/(-18)`

`x = 54/-18`, `2/-18`

`x = -3, -1/9`

Hope this helps!