Solving $\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$ I try to solve this equation:
$$\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$$
So what i did was:
$$x+2+2*\sqrt{x+2}*\sqrt{x-3}+x-3=3x+4$$
$$2*\sqrt{x+2}*\sqrt{x-3}=x+5$$
$$4*{(x+2)}*(x-3)=x^2+25+10x$$
$$4x^2-4x-24=x^2+25+10x$$
$$3x^2-14x-49$$
But this seems to be wrong! What did i wrong?
 A: Note: the original question read $= \sqrt{3x + 5}$ instead of $= \sqrt{3x + 4}$. There is no problem with the fixed question. Maybe just that it's unfinished. The final line should read $3x^2 - 14x + 49 = 0$ rather than just $x^2 - 14x + 49$. After that, solve the quadratic; only one of the solutions ($7$) to the quadratic gives a solution to the original equation (the other ($-7/3$) was introduced by the squaring operations).
Original response: the $3x + 4$ instead of $3x + 5$ on the first line seems a genuine mistake, rather than a typo; you continue with it.
A: first step on the right hand side should be $3x+5$ not $3x+4$
Then proceeding the same way:
$$
\begin{split}
4(x^2-x-6) &= x^2 + 12x + 36 \\
3x^2 - 16x - 60 &= 0
\end{split}
$$
Since the discriminant of that equation is $16^2 - 4 \cdot 3 \cdot 60 < 0$ there are no solutions.
UPDATE you changed the problem now, so you get
$4(x^2-x-6) = x^2 + 10x + 25$ which results in a quadratic $3x^2 - 14x - 49 = 0$ so $(x-7)(3x+7)=0$ and $x=7$ or $x = -7/3$. The second solution is impossible (drives roots to be complex) so $x=7$ is the only one.
A: What makes you think you've done anything wrong? You can factorise
$$3x^2-14x-49 = (3x+7)(x-7)$$
to obtain a solution to your equation.
Beware: one of the solutions from this quadratic is not a solution because it doesn't satisfy the original equation... this often happens when you solve equations by squaring stuff.
P.S. I must have reached the question after the typo was fixed.
