Bring $3x^2-5x+1=x^2+3x-2$ to standard form first, then solve it with the quadratic formula. The roots are:

Move every term to the left: $3x^2-x^2-5x-3x+1+2=0$, i.e. $2x^2-8x+3=0$, so $a=2,b=-8,c=3.x=\frac{8\pm \sqrt{64-24}}{4}=\frac{8\pm \sqrt{40}}{4}=\frac{8\pm 2\sqrt{10}}{4}=\frac{4\pm \sqrt{10}}{2}$. Dividing only the surd by $2a$ leaves $\frac{8\pm \sqrt{10}}{2}$; reading $b$ as +8 gives the negative version; halving twice gives the denominator 4.