|-1+3r|=10

|U| = U when U>=0 and =-U when U<=0.

So we have to consider two cases:

1) -1+3r >= 0, here |-1+3r| = -1+3r = 10. So 3r = 11 and r = 11/3.And we must check whether for this r -1+3r >= 0. This is true (11-1>=0).

2) -1+3r <= 0, |-1+3r| = 1-3r = 10. So 3r = -9 and r = -3.And again for this r -1+3r=-10 < 0.

So both roots are valid: r1 = 11/3 = 3+(2/3), r2 = -3.

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