Question

How do you solve for x in `y = x^2 - 2x`?

Answer 1

Rewrite as `x^2-2x-y=0`. This is a quadratic equation in variable `x`.
Don't be confused, I'm just pointing out that we will temporarily be thinking of `y` as a constant. (a number).

We would solve by factoring if we could, but we can't so we'll use the quadratic formula, which says that
the solutions to `2x^2 + bx +c = 0` are `x=(-b+-sqrt(b^2-4ac))/(2a)`.

In this equation, `a=1`, `b = -2` and `c=-y`, so we get:

`x=(-(-2)+-sqrt((-2)^2-4(1)(-y)))/(2(1))` so

`x= (2+-sqrt(4+4y))/2 =(2+-sqrt(4(1+y)))/2 = (2+-2sqrt(1+y))/2` and finally,

`x=1+-sqrt(1+y)`

(You can check the answer by substituting for `x` and simplifying:

`(1+sqrt(1+y))^2-2(1+sqrt(1+y))` to get `y`.

And similarly for the other solution.