or (y−1)x2+(4y−2)x+3cy−c=0 Now, x is real. Hence, or (2y−1)2−(y−1)(3cy−c)≥0,∀y∈R or ‌‌(4−3c)y2+(−4+c+3c)y+1−c≥0,∀y∈R or ‌‌4−3c>0 and (4c−4)2−4(4−3c)(1−c)≤0 or c<‌
4
3
and 4(c−1)2−(4−3c)(1−c)≤0 or ‌‌c<‌