1. Given equation

We are given

$$2(x+2)^2 f(x)-3(x+2)^2=10 \int_0^x (t+2)f(t)\,dt, \qquad x\ge 0.$$

We need to find $f(2)$.

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2. Differentiate both sides

Let

$$I(x)=\int_0^x (t+2)f(t)\,dt.$$

Then by the Fundamental Theorem of Calculus,

$$I'(x)=(x+2)f(x).$$

Differentiate the given equation with respect to $x$:

$$\frac{d}{dx}\left[2(x+2)^2f(x)-3(x+2)^2\right]=\frac{d}{dx}\left[10I(x)\right].$$

Now,

$$\frac{d}{dx}\left[2(x+2)^2f(x)\right]=4(x+2)f(x)+2(x+2)^2f'(x),$$

and

$$\frac{d}{dx}\left[-3(x+2)^2\right]=-6(x+2).$$

Also,

$$\frac{d}{dx}[10I(x)]=10(x+2)f(x).$$

So,

$$4(x+2)f(x)+2(x+2)^2f'(x)-6(x+2)=10(x+2)f(x).$$

Since $x\ge 0$, we have $x+2>0$, so divide by $2(x+2)$:

$$2f(x)+(x+2)f'(x)-3=5f(x).$$

Thus,

$$(x+2)f'(x)-3f(x)-3=0,$$

or

$$(x+2)f'(x)-3f(x)=3.$$

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3. Solve the differential equation

Rewrite as

$$f'(x)-\frac{3}{x+2}f(x)=\frac{3}{x+2}.$$

This is a linear differential equation.

The integrating factor is

$$\text{I.F.}=e^{\int -\frac{3}{x+2}\,dx}=e^{-3\ln(x+2)}=(x+2)^{-3}.$$

Multiply the equation by $(x+2)^{-3}$:

$$(x+2)^{-3}f'(x)-\frac{3}{x+2}(x+2)^{-3}f(x)=3(x+2)^{-4}.$$

Hence,

$$\frac{d}{dx}\left[f(x)(x+2)^{-3}\right]=3(x+2)^{-4}.$$

Integrate:

$$f(x)(x+2)^{-3}=\int 3(x+2)^{-4}\,dx + C.$$

Now,

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

So,

$$f(x)(x+2)^{-3}=-(x+2)^{-3}+C.$$

Multiply by $(x+2)^3$:

$$f(x)=-1+C(x+2)^3.$$

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4. Find the constant using the original equation at $x=0$

Put $x=0$ in the given equation:

$$2(2)^2 f(0)-3(2)^2=10\int_0^0 (t+2)f(t)dt.$$

Since the integral is $0$,

$$8f(0)-12=0$$

which gives

$$f(0)=\frac{12}{8}=\frac{3}{2}.$$

From

$$f(x)=-1+C(x+2)^3,$$

we get

$$f(0)=-1+8C=\frac{3}{2}.$$

Thus,

$$8C=\frac{5}{2}$$

so

$$C=\frac{5}{16}.$$

Hence,

$$f(x)=-1+\frac{5}{16}(x+2)^3.$$

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5. Compute $f(2)$

$$f(2)=-1+\frac{5}{16}(4)^3.$$

Since $4^3=64$,

$$f(2)=-1+\frac{5}{16}\cdot 64=-1+20=19.$$

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6. Final answer

$$\boxed{19}$$

The derived answer matches the stored correct answer.