How to solve | 2 x + 3 | ≥ - 13 ?

By definition, ${\displaystyle \left|z\right|\ge 0\forall z\in ℝ}$, so, applying this definition to our question, we know that ${\displaystyle |2x+3|\ge 0}$, which is a stronger condition tan ${\displaystyle |2x+3|\ge -13}$ ("stronger" means that ${\displaystyle |2x+3|\ge 0}$ is more restrictive than ${\displaystyle |2x+3|\ge -13}$).
Instead of interpreting the issue as "solve ${\displaystyle |2x+3|\ge -13}$", we are going to read it as "solve ${\displaystyle |2x+3|\ge 0}$" which, in fact, is easier to solve.
In order to solve ${\displaystyle |2x+3|\ge 0}$ we must again remember the definition of ${\displaystyle \left|z\right|}$, which is done by cases:
If ${\displaystyle z\ge 0}$, then ${\displaystyle \left|z\right|=z}$
If ${\displaystyle z<0}$, then ${\displaystyle \left|z\right|=-z}$
Taking this into account for our problem, we can conclude that:
If ${\displaystyle (2x+3)\ge 0\Rightarrow |2x+3|=2x+3}$ and then, ${\displaystyle |2x+3|\ge 0\Rightarrow 2x+3\ge 0\Rightarrow 2x\ge -3\Rightarrow x\ge -\frac{3}{2}}$
If ${\displaystyle (2x+3)<0\Rightarrow |2x+3|=-(2x+3)}$ and then, ${\displaystyle |2x+3|\ge 0\Rightarrow -(2x+3)\ge 0\Rightarrow -2x-3\ge 0\Rightarrow -2x\ge 3\Rightarrow 2x\le -3}$ (observe that the sign of the inequality has changed on changing the sign of both members) ${\displaystyle \Rightarrow x\le -\frac{3}{2}}$
As the result obtained in the first case is ${\displaystyle \forall x\ge -\frac{3}{2}}$ and the result obtained in the second case is ${\displaystyle \forall x\le -\frac{3}{2}}$, both put together give us the final result that the inequation is satisfied ${\displaystyle \forall x\in ℝ}$.