f:R→R, f(x)=3x2+mx+nx2+1.If the range of his function is [−4,3), then find the value of (m + n) is______.

y=3x2+mx+nx2+1or    x2(y−3)−mx+y−n=0    x∈RD≥0⇒ m2−4(y−3)(y−n)≥0or  m2−4y2−ny−3y+3n≥04y2−4y(n+3)+12n−m2≤0             (1)Also given (y+4)(y−3)≤0           y2+y−12≤0                         (2)Comparing (1) and (2), we get 41=−4(n+3)1=12n−m2−12⇒ m=0 and n=−4