If x is real then x2−2x+4x2+2x+4 takes values in the interval

Let y=x2−2x+4x2+2x+4.Then,x2(y−1)+2x(y+1)+4(y−1)=0⇒ 4(y+1)2−16(y−1)2≥0     [∵x is real ]⇒ (y+1)2−{2(y−1)}2≥0⇒ (3−y)(3y−1)≥0⇒1/3≤y≤3