1/7log2(3y-5)=0

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Solution for 1/7log2(3y-5)=0 equation: 


7;8;5;6;0;0

abs(x+2)+abs(x-4) = 0

89.9/18 = 0

Y = 9136.06388859094*11^2-(11*8744.72698391289)-(5082.80794732235*11^3)+1690.18958325029*11^4-(352.684664331863*11^5)+46.5749999970394*11^6-(3.77706679867796*11^7)+0.171527777764804*11^8-(0.00333719135775427*11^9)+3311.99999993904

1;4;5;6;5;0

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