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?

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, $x\ge5$.
