Then, 2xy−5y‌‌=2x2+4x−22 2x2+2x(2−y)−22+5y‌‌=0 Here, x∈R so 4(2−y)2+4×2(22−5y)&≥0 4+y2−4y+44−10y&≥0 y2−14y+48&≥0 (y−6)(y−8)&≥0 This implies, y∈(−∞,6]∪[8,∞) Now, when x is real no value of y lies in (−∞,6]∪[8,∞)