The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$ Solve in positive integers the system of equations:
$$\begin{cases}
2x^2=4y^2+3z^2+2;
\\
13x=4y+3z+29
\end{cases}$$
My work so far:
I used wolframalpha: $x=3,y=1,z=2$. 

$$2x^2=4y^2+3z^2+2\tag1$$
$$13x=4y+3z+29\tag2$$
From $(1)(2)$,
$$13^2(4y^2+3z^2+2)=2(13x)^2=2(4y+3z+29)^2,$$
i.e.
$$644y^2+(-464-48z)y+489z^2-348z-1344=0$$
See this as a quadratic equation on $y$.
Since the discriminant has to be larger than or equal to $0$, we have to have
$$(-464-48z)^2-4\cdot 644(489z^2-348z-1344)\ge 0,$$
i.e.
$$\small -1.5\approx -\frac{236}{155}=\frac{2(87-49\sqrt{81})}{465}\le \frac{2(87-49\sqrt{69})}{465}\le z\le \frac{2(87+49\sqrt{69})}{465}\le \frac{2(87+49\sqrt{81})}{465}=\frac{352}{155}\approx 2.3$$
giving $$z=1,2$$
I think that you can continue from here.
