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: 
$-2x^{3}-x^{2}-6x+9=0$ 

$f(x)=x+x\sqrt{2x+2}$

We are trying to find $f(x)=3$ but notice that for $-1\le x\le 0$ then $f(x)\le 0$.
$f'(x)=1+\sqrt{2x+2}+\frac {x}{\sqrt{2x+2}}>0$ for $x>0$ 
so $f$ is increasing from $f(0)=0$ to $\lim\limits_{x\to+\infty}f(x)=+\infty$
Thus there is an unique $x$ such that $f(x)=3$ and since $f(1)=3$ we are done.
It's not fordidden to have a look at the graph, before trying more complicated stuff.
