Solving the equation $x + \sqrt{2x+1} = 7$ I can't solve 

$x + \sqrt{2x+1} = 7$.

Well, I know the answer is 4, but that is from just reasoning it out. I can't algebraically solve it.
Thus, a step by step is what I really need.
Thanks in advance!!

If $x + \sqrt{2x+1} = 7$, multiply both sides by $x - \sqrt{2x+1}$
to get $x^2 - (2x+1) = 7(x - \sqrt{2x+1})$, or 
$$\sqrt{2x+1} = x - (x^2 - (2x+1))/7 = (-x^2 + 9x+1)/7.$$
Substituting in the original equation,
$7 = x + (-x^2 + 9x+1)/7 = (-x^2 + 16x+1)/7$
or $x^2 - 16x + 48 = 0$.
Solving this, we get $x = 4$ and $x = 12$.
$x = 4$ satisfies the original equation with a positive square root,
and $x = 12$ satisfies it with a negative square root.
You also have to make sure that the 
multiplying by $x - \sqrt{2x+1}$
does introduce any extraneous root(s).
Since doing this throws away the sign of the square root,
this introduces the value $x = 12$.
