Finding Length Of Segment !enter image description here $AB=12$ , $BC=24$ , $CA=20$ $\dfrac{BF}{CG}=\dfrac{3}{5}$ , $\angle FAG=\angle CAG$ Find the length of $FG$. It seems simple but is not easy for me.

let $FG=x,CG=y,AF=z,\dfrac{z}{20}=\dfrac{x}{y}, z^2=20^2+(x+y)^2-2*20*(x+y)cosC <1>$

$z^2-20^2=20^2\dfrac{x^2-y^2}{y^2},cosC=\dfrac{20^2+24^2-12^2}{2*20*24}=\dfrac{13}{15}$

so $<1>$ can be simplify as:

$\dfrac{20^2(x-y)}{y^2}=(x+y)-\dfrac{2*20*13}{15} <2>$

note $BF=\dfrac{3CG}{5}=\dfrac{3y}{5}, BF+FG+CG=24 \implies \dfrac{3y}{5}+x+y=24 \implies y=\dfrac{120-5x}{8}$

$<2>$ becomes: $\dfrac{75*x^3}{512}-\dfrac{2825*x^2}{192}-\dfrac{1575*x}{8}+1575=0$

there are $3$ real roots,but only $x=5.7054$ is available.