3r2+20r+32=0.
Factors of 3=(1,3): factors of 32=(1,32), (2,16), (4,8).
(3r+a)(r+b)=3r2+r(a+3b)+ab, so (a,b)=(1,32), (2,16), (4,8), (32,1), (16,2) or (8,4), and:
a+3b=20:
a b 3b a+3b
1 32 96 97
2 16 48 50
4 8 24 28
8 4 12 20 is the correct combination
16 2 6 22
32 1 3 35
(3r+8)(r+4)=0 so 3r+8=0, 3r=-8, r=-8/3; or r+4=0, r=-4 (two solutions).