For what values of r does the function y = erx satisfy the differential equation y'' − 6y' + 2y = 0? (enter your answers as a comma-separated list.)

You mean y = e^(rx)? Yes, there end up being two values. They should both work. Nobody said the solution had to be unique! 

Use this expression for y to calculated y', then take the derivative again to get y ' ' . 
y' = r e^(rx) y' ' = r^2 e^(rx) 

So you have: r^2 e^(rx) - 6r e^(rx) + 2e^(rx) = 0 
Since e^(rx) can never be 0, we can divide both sides by e^(rx) to get 

r^2 - 6r + 2 = 0  
r1 =(6-√28)/2=3-√ 7 = 0.354 
r2 =(6+√28)/2=3+√ 7 = 5.646

so r = 0.354, 5.646

Is there a picture?