Question 1205062
<pre>
{{{ p=re^2/(r+b)^2}}}

Clear the fraction by multiplying both sides by {{{(r+b)^2}}}

{{{ p(r+b)^2=re^2}}}

Square out {{{(r+b)^2}}} as {{{r^2+2rb+b^2}}}

{{{ p(r^2+2rb+b^2)=re^2}}}

Remove the parentheses by distributing:

{{{ pr^2+2prb+pb^2=re^2}}}

Get the term with {{{r^2}}} first, then terms with r then other terms:

{{{pr^2+2prb-re^2+pb^2=0}}}

Factor r out of the two terms that contain r:

{{{pr^2+(2pb-e^2)r+pb^2=0}}}

Use the quadratic formula: {{{x = (-B +- sqrt( B^2-4AC ))/(2A) }}}
with x=r, A=p, B=(2pb-e<sup>2</sup>) and C=pb^2 

{{{r = (-(2pb-e^2) +- sqrt( (2pb-e^2)^2-4prb^2 ))/(2p) }}}

remove the parentheses of the first term in the numerator
Square the binomial under the square root radical:

{{{r = (-2pb+e^2 +- sqrt(4b^2p^2-4bpe^2+e^4-4b^2p^2 ))/(2p) }}}

The {{{4b^2p^2}}} and {{{-4b^2p^2}}} cancel under the square root radical:

{{{r = (-2pb+e^2 +- sqrt(-4bpe^2+e^4 ))/(2p) }}}

Factor out {{{e^2}}} under the square root radical

{{{r = (-2pb+e^2 +- sqrt(e^2(-4bp+e^2) ))/(2p) }}}

Take out the the square root of {{{e^2}}} as e in front of the square root

{{{r = (-2pb+e^2 +- e*sqrt(-4bp+e^2 ))/(2p) }}}

Reverse the terms under the radical so the positive term comes first
which makes the + sign unnecessary:

{{{r = (-2bp+e^2 +- e*sqrt(e^2 - 4bp))/(2p) }}}

Edwin</pre>