Answer:
[tex]2+\sqrt{6r-5}=r........(1)[/tex]
Step 1: Put the irrational term in left side and the remaining on right side.
[tex]\text{Subtracting 2 from both sides,}\\\\2+\sqrt{6r-5}-2=r-2\\\\\text{or, }\sqrt{6r-5}=r-2[/tex]
Step 2: Square both sides and simplify
[tex](\sqrt{6r-5})^2=(r-2)^2\\\\\text{or, }6r-5=r^2-4r+4[/tex]
[tex]\text{or, }r^2-4r-6r+4+5=0\\\\\text{or, }r^2-10r+9=0[/tex]
Step 3: Factorise left side and find the value of [tex]r.[/tex]
[tex]r^2-9r-r+9=0\\\\\text{or, }r(r-9)-1(r-9)=0\\\\\text{or, }(r-9)(r-1)=0\\\\\text{i.e. }r=1,\ 9[/tex]
Step 4: Put each values of [tex]r[/tex] in your original question to check your answer.
[tex]\text{Put }r=1\text{ in equation(1),}\\\\2+\sqrt{6(1)-5}=1\\\\\text{or, }2+1=1\ (\text{False})[/tex]
[tex]\text{Put }r=9\text{ in equation(1),}\\\\2+\sqrt{6(9)-5}=9\\\\\text{or, }2+\sqrt{49}=9\\\\\text{or, }2+7=9[/tex]
Only the solution [tex]r=9[/tex] satisfies our original equation, so [tex]\boxed{r=9}.[/tex]
Hope this helps!
