Question 1036042
What kind of preceding number?


The number to multiplied by itself, x.
Some unspecified preceding number, n.
{{{x*x=13+n*n}}}
{{{x^2=n^2+13}}}
{{{x=sqrt(n^2+13)}}}, but this should be {{{x=0+- sqrt(n^2+13)}}}.


If your preceding number should be an integer, then you may want {{{n=x-1}}}.  This will give the initial equation as  {{{x^2=(x-1)^2+13}}}  and then you could solve this for x.



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You could accept the assumption I described, and you should be able to <b>solve</b> the resulting {{{x^2=(x-1)^2+13}}} without trouble.