(c) Equation is $|3{x^2} + 12x + 6|\, = 5x + 16$ …..$(i)$

when $3{x^2} + 12x + 6 \ge 0$$ \Leftrightarrow \,\,\,{x^2} + 4x \ge – 2$

$ \Leftrightarrow \,\,\,|x + 2{|^2} \ge 4 – 2\,\,\, \Leftrightarrow \,\,\,|x + 2|\, \ge {(\sqrt 2 )^2}$

$ \Leftrightarrow \,\,\,x + 2 \le – \sqrt 2 $ or $x + 2 \ge \sqrt 2 $ ….$(ii)$

Then $(i)$ becomes $3{x^2} + 12x + 6 = 5x + 16$

$ \Leftrightarrow \,\,\,3{x^2} + 7x – 10 = 0 \Rightarrow x = 1,\, – \frac{{10}}{3}$

But $x = – \frac{{10}}{3}$ does not satisfy $(ii).$

When $3{x^2} + 12x + 6 < 0$$ \Rightarrow {x^2} + 4x < – 2$

== > $\,|x + 2|\, \le \sqrt 2 $$ \Rightarrow \,\, – \sqrt 2 – 2 \le x \le – 2 + \sqrt 2 $ …..$(iii)$

Then $(i)$ becomes $ \Rightarrow 3{x^2} + 12x + 6 = – (5x + 16)$

$ \Rightarrow \,\,\,\,3{x^2} + 17x + 22 = 0 \Rightarrow x = – 2, – \frac{{11}}{3}$

But $x = – \frac{{11}}{3}$does not satisfy $(iii)$. So, $1$ and $-2$ are the only solutions.