How do I find the horizontal tangents of the curve

y=x^3-2x+x+1
All I know is that the slope=0 and I found the derivative:
dy/dx= 3x^2-4x+1

I don't know what to do from here.

Im assuming your equation is actually y=x^3-2x^2+x+1

if so then your derivative is correct

Note to find the horizontal tangents ...you just set this derivative to zero
and solve for x

so we have 3x^2-4x+1=0
(3x - 1)(x - 1 )=0
so we have x=1/3 and x=1
for each of these values we need to compute the y values of the function

for x=1/3 we have y= 31/27
for x=1 we have y=1
so now to the tangent lines them selves

y=31/27 and y=1