Let y=x2−6x+7 ⇒x2−6x+7−y=0
On comparing with ax2+bx+c=0, we get a=1,b=−6 and c=(7−y)
Now, using Sridharacharya formula, x=a−b±b2−4ac =26±36−4(7−y) =26±36−28+4y =26±4y+8 =26±2y+2 =3±y+2 f(x) is defined only when y+2≥0⇒y≥−2 ∴ Range of f(x)=[−2,∞)