Question

How do you solve `3x^2-5x=28`?

Answer 1

`x=-7/3` or `x=4`

Explanation:

We can easily see that the given equation is a quadratic.

First, we subtract 28 from both sides to make the right-hand side equal 0:
`3x^2-5x-28=0`

Now we factor the quadratic and for this we can multiply coefficient of `x^2` and constant term, which is `-84` and identify its two factors whose sum is `-5`, the coefficient of `x`. These numbers are `-12` and `7` and hence `3x^2-5x-28=0` can be written as

`3x^2-12x+7x-28=0`

or `3x(x-4)+7(x-4)=0`

ot `(3x+7)(x-4)=0`

Since either `3x+7` or `x-4` must be equal to 0, we can individually set each one to zero and solve:
`3x+7=0`
`x=-7/3`

`x-4=0`
`x=4`

Therefore, `x=-7/3` or `x=4`