-3√ (2 |2x + 1| - 5) + 6x + 3 = 0

To solve the equation, we can follow these steps:

1. Start by isolating the radical term on one side of the equation:
-3√ (2 |2x + 1| - 5) = -6x - 3

2. Divide both sides of the equation by -3 to get rid of the coefficient in front of the radical:
√ (2 |2x + 1| - 5) = 2x + 1

3. Square both sides of the equation to eliminate the square root:
(√ (2 |2x + 1| - 5))^2 = (2x + 1)^2

4. Simplify both sides of the equation:
2 |2x + 1| - 5 = 4x^2 + 4x + 1

5. Remove the absolute value by considering two cases:
Case 1: 2x + 1 ≥ 0
In this case, the absolute value can be removed:
2(2x + 1) - 5 = 4x^2 + 4x + 1
4x + 2 - 5 = 4x^2 + 4x + 1
4x - 3 = 4x^2 + 4x + 1
4x^2 + 4x + 1 - 4x - 4x - 3 = 0
4x^2 - 3 = 0

Case 2: 2x + 1 < 0
In this case, the absolute value becomes negative:
2(-2x - 1) - 5 = 4x^2 + 4x + 1
-4x - 2 - 5 = 4x^2 + 4x + 1
-4x - 7 = 4x^2 + 4x + 1
4x^2 + 4x + 1 + 4x + 7 = 0
4x^2 + 8x + 8 = 0

6. Solve each quadratic equation separately:
For the first quadratic equation, 4x^2 - 3 = 0:
4x^2 = 3
x^2 = 3/4
x = ±√(3/4)
x = ±√3/2

For the second quadratic equation, 4x^2 + 8x + 8 = 0:
This equation does not have real solutions since the discriminant (b^2 - 4ac) is negative.

Therefore, the solutions to the original equation are:
x = ±√3/2