How do you find the zeros, real and imaginary, of $y=- 5x^2-15x -3$ using the quadratic formula?

Question

1839 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-04T07:03:42

$x = (15 + sqrt(165))/-10$ and $x = (15 - sqrt(165))/-10$

$y = -5x^2 - 15x - 3$

The quadratic formula is $x = (-b +- sqrt(b^2 - 4ac))/(2a)$ .

We know that: $a = -5$ , $b = -15$ , and $c = -3$ based on the equation, so let's plug them into the formula:

$x = (-(-15) +- sqrt((-15)^2 - 4(-5)(-3)))/(2(-5))$

$x = (15 +- sqrt(225 - 60))/-10$

$x = (15 +- sqrt(165))/-10$

Hope this helps!

How do you find the zeros, real and imaginary, of y=- 5x^2-15x -3y=- 5x^2-15x -3 using the quadratic formula?