设x+1/x=t,则有x^2+1/x^2+2=t^2,即有x^2+1/x^2=t^2-2
原方程可以化为:6(t^2-2)+5t=38
6t^2+5t-50=0
(3t+10)(2t-5)=0
t1=-10/3
t2=5/2
故有x+1/x=-10/3
即有x^2+10/3x+1=0
(x+3)(x+1/3)=0
x1=-3,x2=-1/3
x+1/x=5/2
x^2-5/2x+1=0
(x-2)(x-1/2)=0
x3=2,x4=1/2