How do I find the inverse of $y=2x^2+2x+2$

Question

I'm supposed to find the inverse of $$y=2x^2+2x+2 $$ I know that the inverse isn't a function, as it isn't injective, but we're supposed to find it in domain $[0, 1]$. So far, I've switched x for y, and tried to solve for y, and gotten so far: $$ x=2y^2+2y+2$$ $$ 2y^2 + 2y = x-2$$ $$ y(y+1)=1/2 (x-2)$$

WolframAlpha tells me the inverse is supposed to be $$ y= 1/2 (\pm\sqrt{2x-3}-1 ) $$ But I have no idea how to get so far

Answer 1

As suggested in the comments, one should use the quadratic formula. To see this, transform your initial equation in the following:

$$ x = 2y^2 + 2y + 2 \Leftrightarrow 0 = y^2 + y + \left(1-\frac{x}{2}\right).$$

Using the usual quadratic formula in this, we get:

$$y = \frac{-1\pm \sqrt{1-4\left(\frac{2-x}{2}\right)}}{2} = \frac{-1\pm \sqrt{1-4+2x}}{2} = \frac{1}{2}\left( \pm \sqrt{2x-3} - 1\right).$$

Confirming the result your obtained through WolframAlpha.