Question

How do you solve `2x^2 = 3x -2=0`?

Answer 1

I think you meant `2x^2+3x-2 = 0` since `+` and `=` are on the same key.

`0 = 2x^2+3x-2 = (2x-1)(x+2)`

So `x = 1/2` or `x = -2`

Explanation:

Let `f(x) = 2x^2+3x-2`.

If `f(x)` has factors with integer coefficients, then they must be of the form `(2x+-1)(x+-2)` for some combination of `+-` signs.

To see this, first note that `2` only factors as `+-2 xx +- 1`. The coefficient of the `x^2` term is `2`, the constant term is `2` and the middle term is not divisible by `2`. So one of the factors is `(2x+-1)` and the other is `(x+-2)`

Furthermore, the sign of the constant term is `-`, so one of the factors has a `+` and the other a `-`. That leaves only two possibilities to try:

`(2x+1)(x-2) = 2x^2-3x-2`

and

`(2x-1)(x+2) = 2x^2+3x-2`