Answer:Inverse of [tex]\bold{y=x^{2}-10 x}[/tex] is [tex]\bold{y=5 \pm \sqrt{x+25}}[/tex]Solution:Given that[tex]y=x^{2}-10 x[/tex]Adding [tex]5^{2}[/tex] on both sides[tex]y+5^{2}=x^{2}-10 x+5^{2}[/tex]Rewrite 10x as 2(5)x,[tex]y+5^{2}=x^{2}-2(5) x+5^{2}[/tex]By using [tex]\left(a-b)^{2}=a^{2}-2 a b+b^{2}\right[/tex] , we get[tex]y+5^{2}=(x-5)^{2}[/tex][tex]y+25=(x-5)^{2}[/tex] (Completing the square)Now swap x and y, we get[tex]x+25=(y-5)^{2}[/tex]Rewrite the above equation,[tex](y-5)^{2}=x+25[/tex]Taking square root of both sides,[tex]y-5=\pm \sqrt{x+25}[/tex]Adding 5 on both sides,[tex]y=5 \pm \sqrt{x+25}[/tex]Hence inverse of [tex]\bold{y=x^{2}-10 x \text { is } y=5 \pm \sqrt{x+25}}[/tex]