A=πr^2(√h-√r) make "r" the subject of the formula show workings

Answer 1

I think the formula is flawed, since the units on the left and right don't agree. But, going with what we have,
A=πr^2(√h-√r)
Suppose we let z^2 = r. Then we have
A = π√h z^4 - πz^5
degree-5 polynomials cannot be solved algebraically.
Check for a typo.