let \(3 x^{2}+2 x+3=a\)
So \(\frac{a}{1}+\frac{12}{a}=7\)
multiply by a
\[
\begin{aligned}
&a\left(\frac{a}{1}\right)+\left(1 \frac{2}{2}\right) a=7 a \\
&a^{2}-7 a+12=0 \\
&a^{2}-4 a-3 a+12=0 \\
&a(a-4)-3(a-4)=0 \\
&(a-3)(a-4)=0=4
\end{aligned}
\]
\begin{aligned}
&\text { but } a=3 x^{2}+2 x+3 \\
&\therefore 3 x^{2}+2 x=0 \\
&\Rightarrow x(3 x+2)=0 \\
&\Rightarrow x=0 \text { or } x=-2 / 3 \\
&\text { and } \\
&3 x^{2}+2 x+3=4 \\
&3 x^{2}+2 x-1=0 \\
&3 x^{2}+3 x-x-1=0 \\
&(3 x-1)(x+1)=0 \\
&\Rightarrow x=-1 \text { or } x=1 / 3
\end{aligned}