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Simultaneous equations

A-B=14 (1)

AB=120 (2)

transposing(1) A=14+B

substituting into(2)

B(14+B)=120

B^2+14B =120

B^2+14B-120=0

solve quadratic

(B+20)(B-6)=0

B=-20,6

one of the whole numbers is 6 and the other is 6+14=20

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Q: Find two whole numbers that have a difference of 14 and a product of 120?
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