What helps you here is De Morgan's Law, which basically says:
~(a & b) == ~a | ~b
Thus we can just negate this and get:
a & b == ~(~a | ~b) //4 operations
And looking at the truth table (and in fact, god bless the simplicity of binary logic, there are only four possible combintations of inputs to generate the appropriate outputs for) we can see that both are equivalent (last two columns):
a | b | ~a | ~b | ~a OR ~b | ~(~a OR ~b) | a AND b
--|---|----|----|----------|-------------|--------
0 | 0 | 1 | 1 | 1 | 0 | 0
1 | 0 | 0 | 1 | 1 | 0 | 0
0 | 1 | 1 | 0 | 1 | 0 | 0
1 | 1 | 0 | 0 | 0 | 1 | 1