Question

How do you solve `6-7x=3x²`?

Answer 1

`x=2/3, x=-3`

Explanation:

Put all terms onto one side of the equation so that everything is equal to `0`.
`6-7x=3x^2`
`6=3x^2+7x`
`0=3x^2+7x-6`

Factor.
`3x^2+7x-6=0`
`(3x-2)(x+3)=0`

Set each factor equal to `0`. Make sure to do this for both.
`3x-2=0`
`3x=2`
`x=2/3`

`x+3=0`
`x=-3`

Answers: `x=2/3, x=-3`

Answer 2

`2/3 and - 3`

Explanation:

`y = 3x^2 + 7x - 6 = 0`

Use the new transforming method (Socratic Search)
Transformed equation:

`y' = x^2 + 7x - 18 = 0" "rarr (a xx c = 3(-6) = -18)`

Proceed. Find 2 real roots of y', then, divide them by `a = 3.`

Find `2` numbers knowing the sum `(-b = -7))`
and the product `(a xx c = -18)`.

They are: `2, and (-9).`

Consequently, the `2` real roots of `y` are:

`x_1 = 2/a = 2/3`, and `x_2 = -9/3 = -3`

Note. This new method avoids the lengthy factoring by grouping and solving the 2 binomials.