Solution:
Given the inequality:
[tex]x^2+x-12\ge0[/tex]Step 1: Factorize the left-hand side of the inequality.
Thus,
[tex]\begin{gathered} x^2+x-12\ge0 \\ x^2-3x+4x-12\ge0 \\ x(x-3)+4(x-3)\ge0 \\ (x+4)(x-3)\ge0 \end{gathered}[/tex]Step 2: Identify the intervals.
[tex]\begin{gathered} x+4\ge0 \\ \Rightarrow x\le\: -4 \\ x-3\ge0 \\ \Rightarrow\: x\ge\: 3 \end{gathered}[/tex]Hence, the solution to the inequality is
[tex]\: x\le\: -4\text{ or }x\ge\: 3[/tex]
