Note ......we can't have that
a = b −3 and b = a − 3
Rearranging the first, we have that b = a + 3
But b can' t equal a − 3 and a + 3 at the same time !!!
So...I will assume that b = a − 3
a^2 + (a-3)^2 = 15 expand
a^2 + a^2 − 6a + 9 = 15
2a^2 − 6a − 6 = 0
a^2 − 3a − 3 = 0 the only solution to this that makes sense is that a = [ 3 + √21 ] / 2 cm
And b = a − 3 = [ 3 + √21 ] / 2 − 3 = [ − 3 + √21 ] / 2 cm