Finding coefficients in binomials raised to a power Recently was taught the Binomial Theorem and have begun doing practice questions. I have come across a problem which I think is more to do with the gaps in my knowledge, rather than a misunderstanding of the Binomial Theorem.  
Here is the question: The coefficient of $x^3$ in the expansion of $(2+x)(3-ax)^4$ is $30$. Find the values of the constant $a$
My working: $$(3-ax)^4 = \displaystyle\sum_{i=0}^4 \displaystyle \binom{4}{i} 3^{4-i}(ax)^i  $$
$$ =\displaystyle \binom{4}{0}(3)^4+\displaystyle \binom{4}{1}3^3(-ax)^1+\displaystyle \binom{4}{2}3^2(-ax)^2+\displaystyle \binom{4}{3}3(-ax)^3+\displaystyle \binom{4}{4}(-ax)^4$$
$$ = 81 -108ax + 54a^2x^2 - 12a^3x^3 + a^4x^4 $$
$$ (2-x)(81 -108ax + 54a^2x^2 - 12a^3x^3 + a^4x^4)$$ From here I noticed the only way of making an $x^3$ term is to multiply $2$ by $-12a^3x^3$ and to multiply $x$ by $54a^2x^2$ From here, I worked out that $$ -24a^3 + 54a^2 = 30 $$
$$ a^2(-24a+54)=30$$
From here, I worked out that that: $$ a = \pm\sqrt30 $$$$a = 1 $$
From the solutions to this question, I got only 1 correct solution. Which is a = 1. However, my other two solutions are incorrect. What have I done wrong here?? 

$$
\begin{align}
30
&=\left[x^3\right](2+x)(3-ax)^4\\[6pt]
&=2\left[x^3\right](3-ax)^4+\left[x^2\right](3-ax)^4\\
&=2\cdot3^1\cdot(-a)^3\binom{4}{3}+3^2\cdot(-a)^2\binom{4}{2}\\
&=-24a^3+54a^2
\end{align}
$$
Therefore,
$$
\begin{align}
0
&=4a^3-9a^2+5\\
&=(a-1)\left(4a^2-5a-5\right)
\end{align}
$$
Solve for $a$.
