Question

How do you find the roots, real and imaginary, of `y= 8x^2-15x-4-2(3x+3)^2` using the quadratic formula?

Answer 1

`x~~-4.83` or `x~~-0.27`

Explanation:

First multiply out the terms so that the expression can be rearranged into standard form.
`y = 8x^2 - 15x - 4 -2(9x^2+ 18x +9)`
`y = 8x^2 - 18x^2 -15x -36x - 4 - 9`
`y = -10x^2-51x -13`
For a quadratic expression `y=ax^2 + bx+c` the quadratic formula is `x=(-b+-sqrt(b^2-4ac))/(2a`
`x = (-(-51)+-sqrt((-51)^2 -4(-10)(-13)))/(2(-10)`
`x=(51+-sqrt(2601-520))/-20`
`x = -(51+-sqrt(2081))/20`
Because the number under the square root sign is positive, there are no imaginary roots, only real ones.
`x~~-(51+-45.62)/20`
`x~~-4.83` or `x~~-0.27`