You have th following equation;
[tex]3x^2-5x+1=0[/tex]In order to find the solution to the previous equation, use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]In this case, a = 3, b = -5 and c = 1. By replacing these values into the quadratic formula, you obtain:
[tex]\begin{gathered} x=\frac{-(-5)\pm\sqrt[]{(-5)^2-4(3)(1)}}{2(1)} \\ x=\frac{5\pm\sqrt[]{25-12}}{2}=\frac{5\pm\sqrt[]{13}}{2} \\ x=\frac{5\pm3.60}{2}=2.5\pm1.80 \end{gathered}[/tex]Hence, the solutions are:
x = 2.5 + 1.80 = 4.30
x = 2.5 - 1.80 = 0.70
