Solve a simple equation: $x+x\sqrt{(2x+2)}=3$ 
$x+x\sqrt{(2x+2)}=3$

I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did: 
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got: 
$-2x^{3}-x^{2}-6x+9=0$ 

As in the comment $x-1$ is a factor of $f(x) = -2x^3 -x^2-6x+9$. After an easy long division we get $f(x) = (x-1)(-2x^2-3x-9)$. From this we can use the quadratic equation to see the other two roots are not real.
We can do the long division in the following way:
We need to divide $-2x^3-x^2-6x+9$ by $x-1$ so we can ask what times $x-1$ gives the first term $-2x^3$? This is clearly $\boxed{-2x^2}$. But this also gives $+2x^2$ that we didn't want. So we need to take this away from what is left: we now have $-x^2-6x+9 - (+2x^2) = -3x^2-6x +9$. Again we look for a term that when multiplied with $x-1$ gives the highest order term ($-3x^2$). This is $\boxed{-3x}$ but this gives an extra $+3x$. Now we are left with $-6x+9 - (+3x) = -9x+9$. Here we see that this is $\boxed{-9}$ times $(x-1)$. Adding together the terms that we used along the way, we have $-2x^2-3x-9$.
