Question

How do you solve `14n^2 + 24n - 5004 = 0`?

Answer 1

Quadratic formula

Explanation:

The straightforward way is to use the quadratic formula:
`n=(-b+-sqrt(b^2-4ac))/(2a)`
This is for expressions of the form `an^2 + bn + c`, so in your case a=14, b=24, c=-5004 .

A slightly more complicated way is to reverse foil, trying to factor out your a, b, and c so you can write you equation into
`(xn + y)(sn +t) = 0`, where x, y, s, and t are all numbers. Then you can solve one `()` at a time. But if you have a calculator, the quadratic formula is the best way to go. It'll work every time.

Answer 2

`n = -6/7+-15/7sqrt(78)`

Explanation:

Complete the square then use the difference of squares identity to find:

`0 = 7/2(14n^2+24n-5004)`

`color(white)(0) = 49n^2+84n-17514`

`color(white)(0) = (7n)^2+2(7n)(6)+36-17550`

`color(white)(0) = (7n)^2+2(7n)(6)+6^2-(15^2 * 78)`

`color(white)(0) = (7n+6)^2-(15sqrt(78))^2`

`color(white)(0) = ((7n+6)-15sqrt(78))((7n+6)+15sqrt(78))`

`color(white)(0) = (7n+6-15sqrt(78))(7n+6+15sqrt(78))`

Hence:

`7n = -6+-15sqrt(78)`

So:

`n = -6/7+-15/7sqrt(78)`