Solution
[tex]f(r)=r^2-10r+22[/tex]Using
Intermediate Value Theorem,
When we have two points connected by a continuous curve:
one point below the line the other point above the line then there will be at least one place where the curve crosses the line!
[tex]\begin{gathered} \text{ Since, } \\ f(6)=6^2-10(6)+22=-2 \\ \\ f(7)=7^2-10(7)+22=1 \\ \\ \text{ since }f(6)\cdot f(7)<0 \\ \\ \Rightarrow\text{ }\exists\text{ }r\in(6,7)\text{ such that }f(r)=0 \end{gathered}[/tex]