Question

How do you solve `15x^2+7x-55` using the quadratic formula?

Answer 1

`x=(-7+-sqrt(3349))/30`

Explanation:

Strictly speaking, I guess you would like to solve `15x^2+7x-55=0` or in other words find the zeros of `15x^2+7x-55`. You do not "solve" a quadratic expression.

That having been said, the equation `15x^2+7x-55 = 0` is of the form `ax^2+bx+c = 0` with `a=15`, `b=7` and `c=55`.

This has roots (solutions) given by the quadratic formula:

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

`=(-7+-sqrt((-7)^2-(4*15*(-55))))/(2*15)`

`=(-7+-sqrt(49+3300))/30`

`=(-7+-sqrt(3349))/30`

The square root `sqrt(3349)` does not simplify further since the prime factorisation of `3349` is `17*197`, which contains no square factors.