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.
EFGH is an isosceles trapezoid, because a pair of sides are parrelel.
That means Angle F and Angle E are congruent. Furthermore, we can conclude that arc EH and GF are congruent.
x2 - 2x = 56 - 3x
Solve for X, solve for arcs.