Y=3x^2-5x+1 how many points does the graph intersect the x axis

It will touch the x-axis when y=0 so:

3x^2-5x+1=0  subtract 1 from both sides

3x^2-5x=-1  divide all terms by 3

x^2-5x/3=-1/3  halve the linear coefficient, square it, then add that to both sides(5/6)^2=25/36

x^2-5x/3+25/36=25/36-12/36  now the left side is a perfect square

(x-5/6)^2=13/36

(x-5/6)^2=1/3  take the square root of both sides

x-5/6=±√(1/3) add 5/6 to both sides 

x=5/6±√(1/3)

So the graph will intersect the x-axis at two points when:

x=5/6-√(1/3) and 5/6+√(1/3)