Answer:
see below
Step-by-step explanation:
we have
[tex]5x^2-2x+1=0[/tex]
and the quadratic equation is
[tex]ax^2+bx+c=0[/tex]
so we can see that
[tex]a=5\\\\b=-2\\\\c=1[/tex]
and the quadratic formula is
[tex]x=\frac{-b+-\sqrt{b^2-4ac} }{2a}[/tex] it should go this symbol ± instead of +-
so we insert each value
[tex]x=\frac{-(-2)+-\sqrt{(-2)^2-4(5)(1)} }{2(5)} \\\\x=\frac{2+-\sqrt{4-20} }{10}\\\\x=\frac{2+-\sqrt{-16} }{10} \\\\x=\frac{2+-4i}{10} \\\\x_1=\frac{2(1+2i)}{2*5} \\\\x_1=\frac{1+2i}{5} \\\\x_2=\frac{2(1-2i)}{2*5} \\\\x_2=\frac{1-2i}{5}[/tex]
i don't know if you already know about imaginary numbers if you do then the answer is all the work
if you don't know, well just go until
[tex]x=\frac{2+-\sqrt{-16} }{10}[/tex]
and say there is no answer to this, because negatives don't have square roots
