Find the inverses of the following functions

o)$\displaystyle y = 3(x - 1)^2, x \geq 1$

$\displaystyle x = 3(y - 1)^2$

$\displaystyle x = 3(y - 1)(y - 1)$

$\displaystyle x = 3(y^2 - 2y + 1)$

$\displaystyle x = 3y^2 - 6y + 3$

$\displaystyle x - 3 = 3y^2 - 6y$

$\displaystyle x - 3 = 3y(y - 2)$

$\displaystyle \frac {x - 3}{y - 2} = 3y$

$\displaystyle \frac{\frac {x - 3}{y - 2}}{3} = y$

$\displaystyle (3)\frac {x - 3}{y - 2} = y$

$\displaystyle \frac{3x - 9}{y - 2} = y$

$\displaystyle 3x - 9 = \frac {y}{y - 2}$

$\displaystyle 3x - 9 = 1 - \frac {y}{2}$

$\displaystyle 3x - 9 - 1= \frac {y}{2}$

$\displaystyle 3x - 10 = \frac {y}{2}$

$\displaystyle \frac {3x - 10}{2} = y$

I'm way off. . .

And textbook answer is $\displaystyle y = 1 + \sqrt \frac {x}{3}, x \geq 0$