find range of f(x) = x^2 - 6x + 2

For finding the range of the equation we have to find the domain of the inverse function so, 
$$\begin{gathered} y = x^2-6x+2 \\ x^2-6x+2-y=0 \\ Now this equation is a quadratic quation in x so, \\ x = \frac{-(-6)\pm \sqrt{(-6{)}^2-4(2-y)}}{2} \\ =\frac{6\pm \sqrt{36-8+4y}}{2} \\ =3\pm \sqrt{y+7} \\ So domain of above equation will be, \\ y +7 \ge 0 \\ or y \in [ -7 ,\infty ) \\ So range of the original equation x^2-6x+2 is [ -7 ,\infty ). \end{gathered}$$