A curve is such that $\dfrac{d^2y}{dx^2} = 4 - 3x$ and the point $P(2,5)$ lies on the curve. The gradient of the curve at $P$ is 2. Find the equation of the curve.

1. Integrate the second derivative to find the general expression for the first derivative.

$\frac{dy}{dx} = \int (4 - 3x) \, dx$
$\frac{dy}{dx} = 4x - \frac{3}{2}x^2 + c$

2. Use the given gradient at point P to find the constant of integration $c$.

At $P(2, 5)$, $\frac{dy}{dx} = 2$.
$2 = 4(2) - \frac{3}{2}(2)^2 + c$
$2 = 8 - 6 + c$
$c = 0$
So, $\frac{dy}{dx} = 4x - \frac{3}{2}x^2$

3. Integrate the first derivative to find the general equation of the curve.

$y = \int \left(4x - \frac{3}{2}x^2\right) \, dx$
$y = 2x^2 - \frac{1}{2}x^3 + d$

4. Use the coordinates of point P to find the second constant of integration $d$.

At $P(2, 5)$, $y = 5$.
$5 = 2(2)^2 - \frac{1}{2}(2)^3 + d$
$5 = 8 - 4 + d$
$d = 1$
So, $y = 2x^2 - \frac{1}{2}x^3 + 1$