The number of real value of x for which the equality $\left|3{\mathrm{x}}^2+12\mathrm{x}+6\right|=5\mathrm{x}+16$ holds good is

Equation is $\left|3{\mathrm{x}}^2+12\mathrm{x}+6\right|=5\mathrm{x}+16 ...\left(\mathrm{i}\right)$

when $3{\mathrm{x}}^2+12\mathrm{x}+6\ge 0\iff {\mathrm{x}}^2+4\mathrm{x}\ge -2$

$\iff {\left|\mathrm{x}+2\right|}^2\ge 4-2\iff \left|\mathrm{x}+2\right|\ge {\left(\sqrt{2}\right)}^2$

$\iff \mathrm{x}+2\le -\sqrt{2} \mathrm{or} \mathrm{x}+2\ge \sqrt{2} ...\left(\mathrm{ii}\right)$

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

$\iff 3{\mathrm{x}}^2+7\mathrm{x}-10=0\Rightarrow \mathrm{x}=1, -\frac{10}{3}$

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

when $3{\mathrm{x}}^2+12\mathrm{x}+6<0\Rightarrow {\mathrm{x}}^2+4\mathrm{x}<-2$

$\Rightarrow \left|\mathrm{x}+2\right|\le \sqrt{2}\Rightarrow -\sqrt{2}-2\le \mathrm{x}\le -2+\sqrt{2} ...\left(\mathrm{iii}\right)$

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

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

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