Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0 \Rightarrow 12 \cdot 2^{2x} - 35 \cdot 2^x \cdot 3^x + 18 \cdot 3^{2x} = 0 \Rightarrow$
$\Rightarrow 12 \cdot \left(\frac{2}{3}\right)^{2x}-35 \cdot \left(\frac{2}{3}\right)^{x}+18=0 $
Now make substitution : $\left(\frac{2}{3}\right)^{x} = t$ , and solve quadratic equation .
The following substitution may have to work:
$$2^x=t; ~~3^x=s$$
Note that the equation simplifies to, $$12 t^2-35st+ 18s^2=0$$
This factorises to $$(3t-2s)(4t-9s)=0$$
Therefore,
$$3\cdot2^x=2 \cdot 3^x ~~\text{or}~~2^{x+2}=3^{x+2}$$ Since, $(2,3)=1$, we have that $\boxed{x=1~~ \text{or}~~-2}$
Edited to add:
You can view that as a quadratic equation in $t$:
So, the solution will have to be, $$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}$$
This is a bit numerically taxing and honestly, I did not do it this way. Rather, I resorted to something that is equivalent to this. You need to write $18 \times 12$ as product of two numbers whose sum is $35$. Later, prefix a minus sign to these numbers and use them here.
With a little bit of playing around, with factorisations, you'll see they should be $27 \times 8$.
Added by dindoun
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24} =\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24} =\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s$$
$$ \begin{eqnarray} 0 &=& 3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} \\&=& 12 \cdot (2^x)^2 - 35 \cdot (2^x3^x) + 18 \cdot (3^x)^2 \\&=& 12 t^2 - 35 st + 18 s^2 \qquad\text{for}\qquad s=3^x,~t=2^x \\&=& (3t - 2s)(4t - 9s) \\ \implies&& s=\frac32t \quad\text{or}\quad s=\frac49t \\&& s=\left(\frac23\right)^\delta t \quad\text{for} \quad\delta=\frac{1\pm3}{2}=-1~\text{or}~2 \\&& 3^x=\left(\frac23\right)^\delta 2^x \\&& 1=\left(\frac23\right)^{x+\delta} \\&& 0=(x+\delta)\log\left(\frac23\right) \\&& x+\delta=0 \\&& x=-\delta=-\frac{1\pm3}{2}=\boxed{~1~\text{or}~-2~} \end{eqnarray} $$ To make the factorization in the fourth line, from the signs of the coefficients $12,-35,18$, we can see that we need to find a combination of the factors from a row on the left and a row on the right $$ \begin{eqnarray} 1 \cdot 12 &\quad& 1 \cdot 18 \\ 2 \cdot 6 &\quad& 2 \cdot 9 \\ 3 \cdot 4 &\quad& 3 \cdot 6 \end{eqnarray} $$ so that a sum of the products (either of the first numbers and the second numbers, or of the first numbers with the second numbers) is $35$. So taking $3 \cdot 4$ for $12$ and $2 \cdot 9$ for $18$, we can obtain $2\cdot 4+3\cdot 9=8+27=35$ to get the factors above.