2,44=12500x^2/(3+x)^4*(1-x)*(2-2x)^2
2.44=12500x^2/[x^4+12 x^3+54 x^2+108 x+81]*(1 - x)*[4 x^2-8 x+4], solve for x. OK, guys and gals: I have expanded it as much as I could! See if you can perform some magic on it!. Thanks.
+118723 2,44=12500x^2/(3+x)^4*(1-x)*(2-2x)^2
\(2.44=\frac{12500x^2}{(3+x)^4}*(1-x)*(2-2x)^2\\ 2.44=\frac{12500x^2}{(3+x)^4}*(1-x)*4(1-x)^2\\ 2.44=\frac{12500x^2}{(3+x)^4}*4(1-x)^3 \qquad x\ne-3\\ 0.61=\frac{12500x^2}{(3+x)^4}(1-x)^3 \\ 0.61*(3+x)^4=12500x^2(1-x)^3 \\ 0.61*(x^4+4*3x^3+6*9x^2+4*27x+81)=12500x^2(-x^3+3x^2-3x+27) \\ \mbox{ok I'm bored now- got better things to do....} \)
