a)
3x²-5x+m=0,
x=(5±√(25-12m))/6.
25-12m≥0 for non-complex roots.
Therefore 12m≤25, m≤25/12.
EXAMPLE
When m=0, x(3x-5)=0, x=0 or 5/3.
When m=2, x=(5±√(25-24))/6=(5±1)/6=1 or ⅔.
3x²-5x+2=(3x-2)(x-1).
So the solution is m≤25/12 (including, for example, m=0 and m=2).
b)
2(x²+mx)+6=m(x-1),
2x²+2mx-mx+6+m=0,
2x²+mx+6+m=0.
x=(-m±√(m²-8(6+m)))/4,
x=(-m±√(m²-8m-48))/4=(-m±√((m-12)(m+4)))/4.
When m=-4 or 12, the quadratic is a perfect square and has only one distinct root (x=1 when m=-4 and the quadratic is 2x²-4x+2=2(x-1)²; x=-3 when m=12 and the quadratic is 2x²+12x+18=2(x+3)²).
(m-12)(m+4)>0 when m>12 or m<-4.
So the solution to (b) is m>12 or m<-4 (for example, m=14 gives x=-2, -5 and the quadratic is 2x²+14x+20=2(x+2)(x+5); m=-6 gives the quadratic equation 2x²-6x=0=2x(x-3), x=0, 3).