Let $$f(x)$$ and $$g(x)$$ be two polynomials of degree 2 such that $$\frac{f(-2)}{g(-2)}= \frac{f(3)}{g(3)}=4$$.
If $$g(5) =2, f(7)=12,g(7)=-6$$, what is the value of$$f(5)$$?

Let us compare the two quadratics through the difference
$$h(x)=f(x)-4\,g(x)$$

Because $$\frac{f(-2)}{g(-2)}=4$$ and $$\frac{f(3)}{g(3)}=4$$, we have
$$f(-2)=4g(-2) \quad\text{and}\quad f(3)=4g(3)$$

Hence $$h(-2)=0$$ and $$h(3)=0$$. The polynomial $$h(x)$$ is of degree at most $$2$$ and possesses the roots $$x=-2$$ and $$x=3$$, therefore

$$h(x)=k\,(x+2)(x-3)$$ for some constant $$k$$.

Express $$f(x)$$ in terms of $$g(x)$$:
$$f(x)=4\,g(x)+k\,(x+2)(x-3)$$

To determine $$k$$, use the data at $$x=7$$:
$$f(7)=4\,g(7)+k\,(7+2)(7-3)$$

Given $$f(7)=12$$ and $$g(7)=-6$$, substitute:
$$12 = 4(-6) + k\,(9)(4)$$
$$12 = -24 + 36k$$
$$36k = 36$$
$$k = 1$$

Now find $$f(5)$$ using $$g(5)=2$$:

$$f(5)=4\,g(5)+1\,(5+2)(5-3)$$
$$f(5)=4(2)+7\cdot2$$
$$f(5)=8+14$$
$$f(5)=22$$

Therefore, the required value is 22.