Problem:
Let x1<x2<x3 be the three real roots of equation 2014x3−4029x2+2=0. Find x2(x1+x3).
Solution:
Let y=2014. Then the equation becomes 0=yx3−(2y2+1)x2+2=−2x2y2+x3y−(x2−2), a quadratic equation in y. The quadratic formula gives solutions y=x1 or y=2x−x1. In the first case x=y1=20141. In the second case x2−22014x−2=0 which has solutions x=2014±2016. Thus x1=2014−2016,x2=20141, and x3=2014+2016.
Finally, the requested computation is x2(x1+x3)=2.