It is confusing to memorize formulas. To start with, just remember the standard values; freezing and boiling points of water expressed in #C# and #F#. Knowing these data provide the important conversion formula that can be derived from it.

#"Freezing Point of water in " C and F=0^o and 32^o#

#"Boiling Point of water in " C and F=100^o and 212^o#

Now, formulate the ratios that relate the two temperature scales where the temperature units and zero points are different; hence,

#(C-0)/(100-0)=(F-32)/(212-32)#, simplify

#C/100=(F-32)/180#, cross multiply

#100(F-32)=180(C)#, divide both sides by 100 and simplify

#(cancel(100)(F-32))/cancel(100)=(cancel(180)9(C))/(cancel(100)5#, simplify

#F-32=9/5(C)#, add 32 both sides of the equation

#Fcancel(-32+32)=9/5(C)+32#, simplify

#color(red)(F=9/5color(blue)((C))+32-> "the derived formula"#

Then, given the value #color(blue)(C=260^o)#, plug in the value to the formula to convert #C# to #F#; hence

#F=9/5(C)+32#

#F=9/5(260)+32#

#F=468+32#

#F=500^o#