Question 406548
{{{h(x) = x^2 + 12x + 5 = 10}}}  ==> {{{x^2 + 12x - 5 = 0}}} ==> 

{{{x = (-12 +- sqrt( 12^2-4*1*-5))/2  = (-12 +- 2sqrt( 41))/2 

= -6 +- sqrt( 41)

}}} ==> {{{x = -6 + sqrt( 41)}}}

==> {{{h^(-1) (10) = -6 + sqrt( 41)}}}.