Question 105218
I'm going to start where you left off


{{{-3x-4+x^2=0}}} Start with the given equation



{{{x^2-3x-4=0}}} Rearrange the terms



*[invoke quadratic_formula 1, -3, -4, "x"]


However, since {{{x=4}}} makes two terms of the original equation {{{1/(x-4) +x/(x-5) +1/(x^2-9x+20)=0}}} have a denominator of zero, our only true solution is {{{x=-1}}}




So our answer is  {{{x=-1}}}