Question 120070
You wrote:

"{{{b^2+10^2=13^2}}}


I have already tried these setups:


{{{13^2=10^2+b^2}}} and {{{10^2=13^2+b^2}}}"


{{{10^2=13^2+b^2}}} is incorrect: You have the signs wrong.  In order to move the {{{10^2}}} and {{{13^2}}} terms to the opposite sides of the equal sign, you have to subtract them.  The correct equation would be {{{-10^2=-13^2+b^2}}}.  Also neither of your arrangements, even though  {{{13^2=10^2+b^2}}} is technically correct, won't do you much good because you have not isolated the variable in either case.


The arrangement that you want comes from adding {{{-10^2}}} to both sides of the original equation:


{{{b^2+10^2=13^2}}}


{{{b^2+10^2-10^2=13^2-10^2}}}


{{{b^2=13^2-10^2}}}


Notice that we now have the variable, b, isolated to one side of the equal sign and the constant values on the other side.  Now you can take the square root of both sides:


{{{sqrt(b^2)=sqrt(13^2-10^2)}}}


{{{b=sqrt(169-100)=sqrt(69)}}}


So your solution set is {{{b=sqrt(69)}}} or {{{b=-sqrt(69)}}}.  However, if, as I suspect, you are dealing with a right triangle of hypotenuse 13 and long leg of 10 and want the measure of the short leg, you can exclude the {{{b=-sqrt(69)}}} result because a negative number for a length is absurd.  Also {{{sqrt(69)}}} is in simplest terms because there are no perfect square factors of 69 (the prime factorization of 69 is 3 * 23).


Hope this helps.
John