Problem:
Let x1​<x2​<x3​ be the three real roots of equation 2014​x3−4029x2+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​=2014​1​. In the second case x2−22014​x−2=0 which has solutions x=2014​±2016​. Thus x1​=2014​−2016​,x2​=2014​1​, and x3​=2014​+2016​.
Finally, the requested computation is x2​(x1​+x3​)=2​.