Question 860487
{{{S=(v1+v2)^t / 2}}}


Solving for any variable in a formula is a matter of undoing what was done to the variable you want solved.  


Here will be listed only the number properties which justify the steps, but without actually showing those steps:


(!)___Multiplicative Inverse of {{{(1/2)}}};
(!)___Inverse power of t;
(!)___Additive Inverse of v1.


Only one of those reasons might seem confusing.   If you have a expression {{{b^x}}}, and you want to operate on this to have just b, then {{{(b^x)^(1/x)=b^((x)(1/x))=b^1=b}}}.