Question

How do you find the zeros, real and imaginary, of `y=-6x^2-12x+9` using the quadratic formula?

Answer 1

`x=(-2+-sqrt10)/2`

Explanation:

By the quadratic formula,

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

Here, `a=-6`, `b=-12`, `c=9`
So,

`x=(12+-sqrt(144+216))/(-12)`
`rArr` `x=(-2+-sqrt10)/2`

Or in decimal form,

`x=0.58113883008418966599944677221636`
or, `x=-2.5811388300841896659994467722164`