Solution:
$3^{\left | x+2 \right |} + 3^{\left | x-1 \right |} \ge 28$
$\begin{array}{c} \\\\ x+2 \\\\ x-1 \end{array}
\begin{array}{c} (-\infty,-2) \\\\ - \\\\ - \end{array}
\begin{array}{c} [-2,1) \\\\ + \\\\ - \end{array}
\begin{array}{c} [1,+\infty) \\\\ + \\\\ + \end{array}$
Case I. $x \in (-\infty,-2)$
$3^{-(x+2)} + 3^{-(x-1)} \ge 28$
$\frac {1}{3^{(x+2)}} + 3^{1-x} \ge 28$
$\frac {1}{3^{x}\cdot3^{2}} + \frac {3^{1}}{3^{x}} \ge 28$
Let $3^{x}=u$; $u>0$
$\frac {1}{u\cdot9} + \frac {3}{u} \ge 28$
$\frac {1}{u\cdot9} + \frac {3}{u} - 28 \ge 0$
$\frac {1+27-28\cdot9u}{9u} \ge 0 \qquad/ \cdot 9u$
$1+27-252u \ge 0$
$252u \le 28 \qquad / : 28$
$9u \le 1$
$u \le \frac {1}{9}$; but
$u>0 \Leftrightarrow u \in (0,\frac {1}{9}]$
$\begin{array}{|l} u>0 \\\\ u \le \frac {1}{9} \end{array} \Leftrightarrow
\begin{array}{|l} 3^{x}>0 \\\\ 3^{x} \le \frac {1}{9} \end{array} \Leftrightarrow
\begin{array}{|l} \forall x \\\\ 3^{x} \le 3^{-2} \end{array} \Leftrightarrow
x \le -2$
but
Case I. $x \in (-\infty,-2)$
$\Rightarrow x \in (-\infty,-2)$
Case II. $x \in [-2,1)$
$3^{(x+2)} + 3^{-(x-1)} \ge 28$
$3^{(x+2)} + 3^{1-x} \ge 28$
$3^{x}\cdot 3^{2} + \frac {3^{1}}{3^{x}} \ge 28$
Let $3^{x}=u$; $u>0$
$9u + \frac {3}{u} \ge 28$
$9u + \frac {3}{u} - 28 \ge 0$
$\frac {9u^2 + 3 - 28u}{u} \ge 0 \qquad / . u$
$9u^2-28u+3 \ge 0$
$D=676$
$u_{1}=3 > 0$; $u_{2}=\frac {1}{9} > 0$
$9u^2-28u+3 \ge 0 \Rightarrow 9(u-3)(u-\frac {1}{9}) \ge 0 \Leftrightarrow$
$u \in (-\infty,\frac {1}{9}] \cup [3,+\infty)$; but
$u>0 \Rightarrow u \in (0,\frac {1}{9}] \cup [3,+\infty) \Leftrightarrow$
$\begin{array}{|l} u>0 \\\\ u \le \frac {1}{9} \end{array} \Leftrightarrow
\begin{array}{|l} 3^{x}>0 \\\\ 3^{x} \le \frac {1}{9} \end{array} \Leftrightarrow
\begin{array}{|l} \forall x \\\\ 3^{x} \le 3^{-2} \end{array} \Leftrightarrow
x \le -2$
or
$u \ge 3 \Leftrightarrow 3^{x} \ge 3^{1} \Leftrightarrow x \ge 1$
$x \in (-\infty,-2] \cup [1,+\infty)$, but
II. $x \in [-2,1)$
$\Rightarrow x = -2$
Case III. $x \in [1,+\infty)$
$3^{(x+2)} + 3^{(x-1)} \ge 28$
$3^{x}\cdot 3^{2} + 3^{x}\cdot 3^{-1} \ge 28$
$3^{x}\cdot 9 + \frac {3^{x}}{3} \ge 28 \qquad / \cdot 3$
$3\cdot 3^{x}\cdot 9 + 3^{x} \ge 84$
$28\cdot 3^{x} \ge 84$
$3^{x} \ge 3$
$3^{x} \ge 3^{1} \Rightarrow x \ge 1$
Case III. $x \in [1,+\infty)$
$\Rightarrow x \in [1,+\infty)$
Answer: $x \in (-\infty,-2] \cup [1,+\infty) $