⇒‌‌x2y+2xy+3y=x2+14x+9 ⇒‌‌x2(y−1)+2x(y−7)+3y−9=0 Here, x∈R ∴‌‌4(y−7)2−4(y−1)(3y−9)≥0 (y−7)2−(3y2−12y+9)≥0 y2−14y+49−3y2+12y−9≥0 ⇒‌‌2y2+2y−40≤0⇒y2+y−20≤0(y+5)(y−4)≤0‌‌∴‌‌y∈[−5,4] . Maximum and minimum value of ‌