To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = 2(3x + 4)^{\frac{3}{2}} - 6x - 8\).

Integrate each term separately:

1. \(\int 2(3x + 4)^{\frac{3}{2}} \, dx\)

Let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

\(\int 2(3x + 4)^{\frac{3}{2}} \, dx = \frac{2}{3} \int u^{\frac{3}{2}} \, du = \frac{2}{3} \cdot \frac{2}{5} u^{\frac{5}{2}} = \frac{4}{15} (3x + 4)^{\frac{5}{2}}\)

2. \(\int -6x \, dx = -3x^2\)

3. \(\int -8 \, dx = -8x\)

Thus, the integral is:

\(y = \frac{4}{15} (3x + 4)^{\frac{5}{2}} - 3x^2 - 8x + c\)

Given the stationary point \((-1, 5)\), substitute into the equation:

\(5 = \frac{4}{15}(3(-1) + 4)^{\frac{5}{2}} - 3(-1)^2 - 8(-1) + c\)

\(5 = \frac{4}{15}(1)^{\frac{5}{2}} - 3 + 8 + c\)

\(5 = \frac{4}{15} + 5 + c\)

\(c = 5 - 5 - \frac{4}{15} = -\frac{4}{15}\)

Therefore, the equation of the curve is:

\(y = \frac{4}{15}(3x + 4)^{\frac{5}{2}} - 3x^2 - 8x - \frac{4}{15}\)