I have 2 problems I cannot solve :

1) (3m-2)x^2 +2mx+3m=0

Can we determine values of m for which this equation has 2 real roots x' and x" such as 0<x'<1<x" ?

2) x^2 -2(2m-5)x+29-20m=0 (1)

a) x' and x" being the roots of the equation, calculate, as a function of m the expression :

y=1/(x'-3) + 1/(x"-3)

b)can we determine m so that y=1 ? So that y=-4/17 ?

c) Find a relation between x' and x" indenpendent of m. Find using this relation the values of the "double roots" (that is when x'=x") of the equation (1).