Solve for r when given A = 2πr^2+2πr

Question

This is identical to solving for x in the standard form of a quadratic equation ax²+bx+c=0 only that in this case, a=2π, b=2π and c=A, and x=r. Consider substituting in these values and then solving. Your answer should be identical to the quadratic formula before substituting in the alternate values for a, b, c, and x respectively.

Answer 1

2πr² + 2πr - A = 0

r = [-2π ± √(4π² + 8πA)] / (4π)

Answer 2

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