Answer:
[tex]x=\frac{7+\sqrt{29}}{10},\:x=\frac{7-\sqrt{29}}{10}[/tex]
Step-by-step explanation:
[tex]5x^2-7x+1=0[/tex]
Solving with the quadratic formula:
[tex]x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:5\cdot \:1}}{2\cdot \:5}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{29}}{2\cdot \:5}[/tex]
[tex]x_1=\frac{-\left(-7\right)+\sqrt{29}}{2\cdot \:5},\:x_2=\frac{-\left(-7\right)-\sqrt{29}}{2\cdot \:5}[/tex]
[tex]x=\frac{7+\sqrt{29}}{10},\:x=\frac{7-\sqrt{29}}{10}[/tex]
