Quadratic equation with different indices My maths teacher gave me a worksheet to work through as I was getting slightly bored in lessons. However, there was one question which I cannot do. The worksheet gives the answer, but you are supposed to show how you did it. Here is the question:  
$729 + 3^{2x+1} = 4\times3^{x+2}$
The sheet said that the answer was $x=2$ or $x=3$
Here is my working - I 'solved' by completing the square, as you will see:
\begin{align}
 3^{2x+1}-4\times3^{x+2}+729& = 0 \\
 3(3^{2x})-4\times(3^2)(3^x)+729& = 0\\ 
 3(3^x)^2-36(3^x)+729& = 0\\ 
 y  &= 3^x\\
 3y^2-36y+729 & = 0\\
 3(y^2-12y+243) &= 0\\
 (y-6)^2-36+243 &= 0\\
 (y-6)^2+207 &= 0\\
 (y-6)^2&=-207\\
 y-6 &= \sqrt{207}i\\
 y &= 6+\sqrt{207}i\\
 3^x &= 6+\sqrt{207}i\\
 x &= \log_3(6+\sqrt{207}i)\\
\end{align} 
Clearly, this is not the answer as stated on the sheet. Have I answered this question wrong, or is the answer on the sheet wrong? Thank you in advance.

HINT
Set $3^x=y$
$$729 + 3y^2= 4\cdot 9 y$$
$$3y^2-36y+729=0$$

$$y^2-12y+243=0$$

$\Delta <0 \implies$ no real solution 
