To find the equation of the tangent, we first need to find the derivative \(\frac{dy}{dx}\) of the curve.

The given equation is \(y = 2x^{\frac{3}{2}} - 3x - 4x^{\frac{1}{2}} + 4\).

Differentiate term by term:

\(\frac{dy}{dx} = \frac{d}{dx}(2x^{\frac{3}{2}}) - \frac{d}{dx}(3x) - \frac{d}{dx}(4x^{\frac{1}{2}}) + \frac{d}{dx}(4)\).

\(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3 - 2x^{-\frac{1}{2}}\).

Now, evaluate \(\frac{dy}{dx}\) at \(x = 4\):

\(\frac{dy}{dx} \bigg|_{x=4} = 3(4)^{\frac{1}{2}} - 3 - 2(4)^{-\frac{1}{2}}\).

\(\frac{dy}{dx} \bigg|_{x=4} = 6 - 3 - 1 = 2\).

The gradient of the tangent at \(x = 4\) is 2.

The point of tangency is (4, 0). Using the point-slope form of the equation of a line:

\(y - y_1 = m(x - x_1)\), where \(m = 2\), \(x_1 = 4\), and \(y_1 = 0\).

\(y - 0 = 2(x - 4)\).

Thus, the equation of the tangent is \(y = 2(x - 4)\).