quadratic equation what am I doing wrong? solve
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} $$
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} \Rightarrow $$ 
$$ 5x+19 = (x+7) + 4\sqrt{x-5}\sqrt{x+7} + (x+5) \Rightarrow $$ 
$$ 3x + 17 = 4\sqrt{x-5}\sqrt{x+7} \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x+7)(x-5) \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x^2+ 2x - 35) \Rightarrow $$
$$ 7x^2 - 70x - 849 = 0 \Rightarrow $$
$$ b^2 - 4ac = (-70)^2 - 4 \cdot 7 \cdot (-849) = 28672=2^{12}\cdot7 $$
then I calculate the solution using the discriminant as
$$ 5 + 32\frac{\sqrt7}7 $$
and
$$ 5 - 32\frac{\sqrt7}7 $$
but when I plug in the values I find out that they are wrong,
does it have to do with the fact that I square the equation twice?
if so what is the best way to go about solving this equation?
 A: You forgot a factor $4$ and wrote $x+5$ instead of $x-5$ in the third line, which should be
$$
5x+19=x+7+4\sqrt{x+7}\,\sqrt{x-5}+4(x-5)
$$
giving
$$
4\sqrt{x+7}\,\sqrt{x-5}=32
$$
or
$$
\sqrt{x+7}\,\sqrt{x-5}=8
$$
that becomes, after squaring,
$$
x^2+2x-99=0
$$
The roots of this are $-11$ and $9$, but only the latter is a solution of the original equation, because the existence conditions on the radicals give
\begin{cases}
5x+19\ge0\\[3px]
x+7\ge0\\[3px]
x-5\ge0
\end{cases}
that is, $x\ge5$.
A: \begin{align}
\sqrt{5x+19}&=\sqrt{x+7}+2\sqrt{x-5}\\
5x+19&=x+7+4(x-5)+4\sqrt{(x+7)(x-5)}\\
32&=4\sqrt{(x+7)(x-5)}\\
8&=\sqrt{(x+7)(x-5)}\\
64&=x^2+2x-35\\
0&=x^2+2x-99
\end{align}
which gives $x=9$ and $x=-11$ are the solutions. 
But both are giving something like this..
$x=9\implies \sqrt {5(9)+19}=\sqrt{9+7}+2\sqrt{9-5}$
$\implies\sqrt{64}=\sqrt{16}+2\sqrt{4}\implies 8=8$ correct know...!!!!
$x=-11\implies\sqrt {5(-11)+19}=\sqrt{-11+7}+2\sqrt{-11-5}$
$\implies\sqrt{-36}=\sqrt{-4}+2\sqrt{-16}\implies 6i=2i+4i$
But by the series of comments given by the well wishers.. i could understand $x=-11$ is not possible...
