jmc

We can write the given equation as $$\sqrt{3x^2-8x+1}+\sqrt{3(3x^2-8x+1)-11}=3.$$Thus, let $y=3x^2-8x+1,$ so $$\sqrt{y}+\sqrt{3y-11}=3.$$Then $\sqrt{3y-11}=3-\sqrt{y}.$ Squaring both sides, we get $$3y-11=9-6\sqrt{y}+y.$$Then $20-2y=6\sqrt{y},$ so $10-y=3\sqrt{y}.$ Squaring both sides, we get $$y^2-20y+100=9y.$$Then $y^2-29y+100=0,$ which factors as $(y-4)(y-25)=0.$ Thus, $y=4$ or $y=25.$ But only $y=4$ satisfies $10-y=3\sqrt{y}.$

Then $3x^2-8x+1=4,$ so $$3x^2-8x-3=0.$$This leads to the solutions $\boxed{3,-\frac{1}{3}}.$ We check that these solutions work.