We have
$$f(x)=ax^2+bx+c = a(x+3)^2+b(x+3)+c = a(x^2+6x+9)+bx+3b + c = ax^2+6ax+9a + bx+3b + c = ax^2+(6a+b)x+(9a+3b+c)$$
Since $f(x+3)=3x^2+7x+4 = ax^2+(6a+b)+(9a+3b+c)$ we obtain the following system of equations
$$\left \{ \begin{array}{lll} a = 3 \\ 6a+b = 7 \\ 9a+3b+c = 4 \end{array} \right. \Rightarrow \left \{ \begin{array}{lll} a = 3 \\b =-11 \\ c c = 10 \end{array} \right.$$