Question

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

Answer 1

`x = +1 1/2 color(white)(.........)x=-1`

`{EEx : AA x in RR}color(white)(....)` Calculating y is just a matter of substituting for `x`. However it is obvious that `y=0`

Explanation:

Given:`color(white)(..)y=-2x^2+x+3`

Standard form equation: `color(white)(..)y=ax^2+bx+c`

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

In this case:

`a=(-2)`
`b=1`
`c=3`

So`color(white)(....) x= (-1+-sqrt(1^2-4(-2)(3)))/(2(-2))`

`color(white)(....) x= (-1+-sqrt(1+24))/(-4)`

`x=(-1+-5)/(-4)`

`x = +1 1/2 color(white)(.........)x=-1`