3 Ca3 (PO4)2 + 4 CaO
So we can see that there are 3 Ca3 (Ca3, Ca3, Ca3), which means you multiply the subscript by the coefficient, meaning 9 Ca. Then we see (PO4)2, which means that there are two PO4 (PO4, PO4) so we multiply the subscript 2 by the subscript 4 on O to equal 8 O, and we multiply the subscript 2 by the subscript 1 on P (the 1 is not written) to equal 2 P.
So far we have 9 Ca, 8 O, and 2 P.
One last term, 4 CaO (means CaO, CaO, CaO, CaO) which translates to 4 Ca and 4 O.
Add them all together and you get (9+4=13) 13 Ca, (8+4=12) 12 O, and 2 P
