$$ 9^a + 243 =28\cdot 3^{a+1} $$
Determine the values $a$ can take.
Let us recall that $$3^a =t $$
Then we have
$$t^2 + 3^5 =28\cdot 3t$$
Dividing the both sides by $28$
$$\dfrac{t^2 + 3^5}{28}=3t$$
Here we get two solutions
$$a =\{81,3\}$$
However, I believe that I have gone wrong somewhere.
We have $$9^a+243=28\cdot3^{a+1}\implies (3^a)^2-28\cdot3\cdot3^a+243=0$$ and solve the quadratic for $3^a$ to get $$3^a=81,3.$$ You forgot to do $\log_3$ on both sides.
Your final equation is right but your solution for it is wrong$$t=\dfrac{42\pm\sqrt{42^2-4\cdot 243}}{2}=14\pm 3\sqrt {22}$$where only $14+ 3\sqrt {22}$ is valid ($14-3\sqrt {22}<0$) so we have $$x=\log_3{14+ 3\sqrt {22}}$$
$$3^{2a}+3^5=28\cdot 3\cdot3^a$$
$$3^{2a}-84(3^a)+243=0$$ $$y=3^a\to y^2-84y+243=0$$ $$y=81, y=3$$ $$a=\log_3{(y)}\to a=4,1$$
