An average human weighs about 700 N .If two such generic humans each carried 3.0 C coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 700 N weight?

An average human weighs about 700 N .If two such generic humans each carried 3.0 C coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 700 N weight?

So first I AM BLIND! YOUR IMAGES/EQUATION EDITOR CAN'T BE READ BY MY SCREEN READER!! SO FOR HELP PLEASE JUST WRITE IT OUT IN HTML TEXT (THE LITERL CHARACTERS YOU ARE READING RIGHT NOW) PLEASE!! So I know f=1/(4pi*epsilon)*abs(q1*q2))/r^(2)        and we are looking for R here. o first epsilon = 9.8*10^(9) and with all of that information you need to manipulate to solve for R. I get r= sqrt(1/4*pi()*epilison)*abs(q1*q2). From there I am getting a wrong answer and can't figure out what is going on with thealgebra. So could someone help me figur out what needs to be done to the initial equation to solve for r? and finally again please no graphs, images, or anything and  just  can read the html text. appreciate the help if someone  is willing to help and not just reject this for some stupid excuse. :)