If a and b are the numbers, ab=30 and a+b=7.
Substitute b=7-a in ab=30: a(7-a)=30, from which a^2-7a+30=0 and a=(7±sqrt(49-120))/2=(7±i*sqrt(71))/2.
The roots are complex because the sum of the numbers is too small compared to their product to remain real.
This solution for a will apply to b also, because the equations are symmetrical for a and b.
The two numbers are (7+isqrt(71))/2 and (7-isqrt(71))/2.