^{2}+11x+k=0 to have two (different) solutions, it should be 121-4k>0. From here it follows that k<121/4. Since it is required k to be a positive integer, 0<kâ¤30=[121/4], (here [a] denotes the integer part of a)

_{1}=(1/2)(-11+â(121-4k)) and x

_{2}=(1/2)(-11-â(121-4k)). Next, we have to select those k's for which x

_{1}and x

_{2}are integers. For k=10, 18, 24, 28 and 30 (which sums up exactly to 110 as you mentioned above) x

_{1}and x

_{2}are Integers. The corresponding pair of solutions (roots) are: {-1, -10}, {-2, -9}, {-3, -8}, {-4, -7} , {-5, -6}

_{1}and x

_{2}.

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