Question

How do you solve and factor `3x^2+4x-224=0`?

Answer 1

`3x^2+4x-224 = 0`
is a quadratic of the form
`ax^2 + bx + c = 0`
which is solved using the formula
`x = (-b +- sqrt(b^2 - 4ac))/(2a)`
I would recommend chanting this formula as many times as necessary to remember it forever; this formula will keep showing up.

For the given equation we have
`x = (-4 +-sqrt(16+2688))/(6)`

`x = (-4 +- 52)/(6)`

So (after simplifying) the solutions to the given equations are
`x = -(28)/3` and `x = 8`

Therefore 2 of the factors of the quadratic are
`(x+(28)/3)` and `(x-8)`
(since setting `x` to one of our solutions will cause the result to be `0` as required.

However
`(x+(28)/3)(x-8)` is only equal to `1/3` of the original quadratic
so the complete factorization should be
`(3)(x+(28)/3)(x-8)`

which would normally be rewritten to clear the fraction as
`(3x+28)(x-8)`