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(C) The given equation is $|3x^2 + 12x + 6| = 5x + 16$ ... $(i)$Case $1$: $3x^2 + 12x + 6 \ge 0$$x^2 + 4x + 2 \ge 0$ $\Rightarrow (x+2)^2 \ge 2$ $\Rig

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$2$

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(C) The given equation is $|3x^2 + 12x + 6| = 5x + 16$ ... $(i)$Case $1$: $3x^2 + 12x + 6 \ge 0$$x^2 + 4x + 2 \ge 0$ $\Rightarrow (x+2)^2 \ge 2$ $\Rightarrow x \le -2 - \sqrt{2}$ or $x \ge -2 + \sqrt{2}$ ... $(ii)$In this case,the equation becomes $3x^2 + 12x + 6 = 5x + 16 \Rightarrow 3x^2 + 7x - 10 = 0$.Solving the quadratic,we get $(3x + 10)(x - 1) = 0$,so $x = 1$ or $x = -\frac{10}{3}$.Checking against $(ii)$,$x = 1$ satisfies the condition,but $x = -\frac{10}{3} \approx -3.33$ does not satisfy $x \le -2 - 1.414 = -3.414$ (since $-3.33 > -3.414$).Case $2$: $3x^2 + 12x + 6 $-2 - \sqrt{2} In this case,the equation becomes $-(3x^2 + 12x + 6) = 5x + 16 \Rightarrow 3x^2 + 17x + 22 = 0$.Solving the quadratic,we get $(3x + 11)(x + 2) = 0$,so $x = -2$ or $x = -\frac{11}{3}$.Checking against $(iii)$,$x = -2$ satisfies the condition,but $x = -\frac{11}{3} \approx -3.66$ does not satisfy the condition (since $-3.66 Thus,the valid solutions are $x = 1$ and $x = -2$. There are $2$ such real values.

The number of real values of $x$ for which the equality $|3x^2 + 12x + 6| = 5x + 16$ holds is:

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