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In the subtraction I am going to leave off the base 14 but the subtraction is done in base 14. We have
Starting with the units column C is larger than 3 so you need to borrow from the fourteens column. Fourteen plus 3 is 17 so in the units column you have seventeen minus twelve (C) which is 5. Thus you have
Notice that the 1 in the fourteens column has been reduced to zero by the borrowing. Again in the fourteens column you need to subtract from the fourteens squared column. Fourteen plus zero is fourteen and fourteen minus two is twelve which is C in base 14. Thus at this point you have
Notice again that the C is the fourteens squared column has ben educed to B by he borrowing. Now the fourteens squared column. Again you need to borrow and B + fourteen is eleven plus fourteen or twentyfive. Twentyfive minus twelve (C) is thirteen which is D in base 14 so now you are at
Finally you have one minus zero which is one so
Thus $2C13_{14}  C2C_{14} = 1DC5_{14}.$ You can check the answer by addition, that is show that $1DC5_{14} + C2C_{14} = 2C13_{14}.$ Now try your problem and write back if you need more assistance,  


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