First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = 2\sqrt{3x+4} - x\).

Using the chain rule, \(\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1\).

Substitute \(x = 4\) into \(\frac{dy}{dx}\) to find the gradient of the tangent at the point (4, 4):

\(\frac{dy}{dx} = \frac{1}{4}\).

The gradient of the normal is the negative reciprocal of the gradient of the tangent:

\(m = -\frac{1}{\frac{1}{4}} = 4\).

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), with \(m = 4\) and point (4, 4):

\(y - 4 = 4(x - 4)\).

Simplify to find the equation of the normal:

\(y = 4x - 16 + 4\).

Thus, \(y = 4x - 12\).