The solution set of |2x + 3| > 5 is:

Given: |2x + 3| > 5

Let 2x + 3 > 5 and 2x + 3 < - 5 (by the property of modulus)

So, first take 2x + 3 > 5

2x + 3 - 3 > 5 - 3

2x > 2

x > 1

Hence, x $\mathbf{\in }$ (1, $\mathbf{\infty }$)

Now take 2x + 3 < - 5

2x + 3 - 3 < -5 - 3

2x < - 8

x＜ -4

Hence, x $\mathbf{\in }$ ($\mathbf{-}\mathbf{\infty }$, - 4)

 x $\mathbf{\in }$ ($\mathbf{-}\mathbf{\infty }$, - 4) $\mathbf{\cup }$ (1, $\mathbf{\infty }$)

So, the correct answer for |2x + 3| > 5 is (-$\mathbf{\infty }$, -4) $\mathbf{\cup }$ (1, $\mathbf{\infty }$).