Question

How do you find the roots, real and imaginary, of `y= -2x^2 + 15x + 22` using the quadratic formula?

Answer 1

at `y=0` `color(white)(.........)x= + 8.756 " or " -1.256`

Explanation:

Given: `color(white)(...)color(brown)(y=-2x^2+15x+22)`

Using standard for of `y=ax^2+bx+c=0`

Where: `color(white)(....)x=(-b+-sqrt(b^2-4ac))/(2a)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`a=-2`
`b=15`
`c=22`

Thus: `color(white)(.....)x=(-15+-sqrt(15^2-4(-2)(22)))/(2(-2))`

`x= (-15+-sqrt(225 + 176))/(-4)`

`x=(-15+-sqrt(401))/(-4)`

But 401 is a prime number so unable to break it down further

Using `sqrt(401)=20.025` to 3 decimal places

`x= + 8.756 " or " -1.256`