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!!
 A: There are several ways of doing this. Here's one: set $u=\sqrt{2x+1}$. Then $u^2 = 2x+1$, so $2x = u^2-1$, $x=\frac{1}{2}(u^2-1)$. Substituting into the original equation, you have
$$\frac{1}{2}(u^2-1) + u = 7$$
which is a quadratic equation in $u$:
$$u^2 +2u -15 = 0.$$
We can solve this equation for $u$. Then solve $u=\sqrt{2x+1}$ for $x$, remembering that $u$ must be nonnegative, and $x$ must be greater than or equal to $-\frac{1}{2}$. Plug into the original equation to verify they give you correct answers. 
Another method (essentially the same, but without introducing new variables): Rewrite as
$$\sqrt{2x+1} = 7-x.$$
Square both sides, solve for $x$, then plug back into the original equation to verify (squaring both sides may introduce "extraneous solutions", so you need to check the answers you got actually solve the original). 
A: $7-x= \sqrt{2x+1}$
$ (7-x)^2 = 2x+1$
$ x^2-14x+49 = 2x+1 $
$x^2-16x+48=0$
$(x-8)^2-64+48=0$
$(x-8)^2=16$
$x-8=\pm 4$
$x=12$ or $x=4$
But $x=12$ does not work in the original equation. So the answer is $x=4$.
(Or the original equation requires $7-x\ge 0$ and so $x\le 7$.)
A: $x+\sqrt{2x+1}=7$
$(x-7)^2=2x+1$
$x^2-14x+49-2x-1=0$
$x^2-16x+48=0$
$(x-12)(x-4)=0$
$x=12,\ x=4$
$S={4}$
A: 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 $x = 12$.
