Find the solutions of the quadratic equation 3x^2-5x+1=0.

Answer:The solutions of the quadratic equation are [tex]x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}[/tex]Step-by-step explanation:This is a second order polynomial, and we can find it's roots by the Bhaskara formula.Explanation of the bhaskara formula:Given a second order polynomial expressed by the following equation:[tex]ax^{2} + bx + c, a\neq0[/tex].This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex][tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex][tex]\bigtriangleup = b^{2} - 4ac[/tex]For this problem, we have to find [tex]x_{1} \text{and} x_{2}[/tex].The polynomial is [tex]3x^{2} - 5x +1[/tex], so a = 3, b = -5, c = 1.Solution[tex]\bigtriangleup = b^{2} - 4ac = (-5)^{2} - 4*3*1 = 25 - 12 = 13[/tex][tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) + \sqrt{13}}{2*3} = \frac{5 + \sqrt{13}}{6}[/tex][tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) - \sqrt{13}}{2*3} = \frac{5 - \sqrt{13}}{6}[/tex]The solutions of the quadratic equation are [tex]x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}[/tex]