Since, x is real. Then, its discriminant should be ≥0. ∴‌4(y+3)2−4(y−1)(y−5)≥0 ⇒(y+3)2−(y−1)(y−5)≥0 ⇒y2+9+6y−(y2−6y+5)≥0 ⇒y2+9+6y−y2+6y−5≥0 ⇒12y+4≥0 ⇒4(3y+1)≥0 ⇒y≥−1∕3 So, least value of given expression is (‌