Trapezoid $EFGH$ is inscribed in a circle, with $EF \parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees.

220^{o} since arc(HE) and arc(GF) are both 35^{o} and arc(HG) is 70^{o} so arc EPF is (360-35-35-70)^{o}=220^{o} and we have that arc(EPF) is \(\boxed{220^\circ}.\)