$\displaystyle
\frac{(2x-14)+12(\sqrt{x+2})}{\sqrt{x+2}}=0
$
From the domain $\displaystyle x > -2$
$\displaystyle (2x-14) + 12\sqrt{x+2} = 0$
$\displaystyle 2(x-7) = -12 \sqrt{x+2}$
Cancel out 2 then square:
$\displaystyle x^2-14x+49 = 36x+72$
$\displaystyle x^2-50x-23 = 0$
23 is prime and so solve using quadratic formula
$\displaystyle x = \frac{50 \pm \sqrt{50^2-4 \cdot 1 \cdot -23}}{2}$
$\displaystyle x= 25 \pm 18 \sqrt{2}$
However, if we put these solutions back into the original equation we find that $\displaystyle 25 + 18\sqrt{2} \neq 0$ - it is an extraneous root so it should be discarded
$\displaystyle x= 25 - 18 \sqrt{2}$ does equal 0 when put into the original equation so this is a root
The numerator must be zero, so ignore the denominator, except to see limitations of acceptable values [x NOT<=-2 or it is undefined or complex.]
Divide by 2 to simplify.
Move the radical to one side [watch for sign change] and square both sides.
Give it a try.