[tex]\begin{gathered} y=x^2-10x \\ \Rightarrow x^2-10x-y=0 \end{gathered}[/tex][tex]\Rightarrow(x-5)^2-(-5)^2-y=0[/tex][tex]\begin{gathered} \Rightarrow(x-5)^2=y+25 \\ \Rightarrow x-5=\pm\sqrt[]{y+25} \end{gathered}[/tex][tex]\Rightarrow x=\pm\sqrt[]{y+25}+5[/tex]
Hence the inverse is given by:
[tex]y=\pm\sqrt[]{x+25}+5[/tex]
