Question 987650
{{{A = 2*pi*r^2 + 2*pi*r*h}}}



{{{A-A = 2*pi*r^2 + 2*pi*r*h-A}}}



{{{0 = 2*pi*r^2 + 2*pi*r*h-A}}}



{{{2*pi*r^2 + 2*pi*r*h-A = 0}}}



{{{2*pi*r^2 + 2*pi*h*r-A = 0}}}



The last equation is in the form {{{ax^2+bx+c=0}}} where {{{a = 2pi}}}, {{{b = 2pi*h}}} and {{{c = -A}}}



Use the quadratic formula to solve for r



{{{r = (-b+-sqrt(b^2-4ac))/(2a)}}}



{{{r = (-2pi*h+-sqrt((2pi*h)^2-4*2pi*(-A)))/(2(2pi))}}}



{{{r = (-2pi*h+-sqrt(4pi^2*h^2+8Api))/(4pi)}}}



Since {{{r > 0}}}, this means we only focus on the "plus" portion of the "plus/minus". So we end up with



{{{r = (-2pi*h+sqrt(4pi^2*h^2+8Api))/(4pi)}}}