Quadratic formula tells us that the solution of quadratic equation #ax^2+bx+c=0# is given by
#x=(-b+-sqrt(b^2-4ac))/(2a)#, where #a# is coefficient of #x^2#, #b# is coefficient of #x# and #c# is constant term. This is known as quadratic formula.
As #A=2pir^2+2pirh# is a quadratic equation in #r# and we can use quadratic formula to get solution of the equation.
This means you want to find the radius of a right cylindrical solid, whose total surface area #A# and height #h# are known. In that case we proceed as follows.
We can write the equation as
#2pir^2+2pihr-A=0#
Here we have a quadratic equation in #r# and coefficient of #r^2# is #2pi#, coefficient of #r# is #2pih#and constant term is #-A#.
Hence #r=(-2pih+-sqrt((2pih)^2-4*(2pi)*(-A)))/(2*2pi)#
= #(-2pih+-sqrt(4pi^2h^2+8piA))/(4pi)#
= #(-2pih+-2sqrt(pi^2h^2+2piA))/(4pi)#
= #(-pih+-sqrt(pi^2h^2+2piA))/(2pi)#
But here #r# can only be positive, so we must have #+# sign only as we already have #-pih#.
Hence #r=(sqrt(pi^2h^2+2piA)-pih)/(2pi)#
Observe that #sqrt(pi^2h^2+2piA)>pih#, hence we will have a positive solution to #r#,
