Поясніть, як це робити!
объясните как это делать!

Пусть 2 выполнит всю работу за x ч, тогда 1 за x+5 ч. вся работа - 1 => производительность 1 - (1/x+5); 2 - (1/x). всего совместной работы 8ч, но 1 на 2 часа больше(раньше начал). составим уравнение:
10*(1/x+5) + 8*(1/x) = 0.8 (вместе) //домножим на 10
100/(x+5) + 80/x=8 //перенесем 8 <-, 80/x->
100/(x+5) - 8 =-80/x
100-8*x-40/(x+5)=-80/x
-8*x+60/(x+5)=-80/x
-8*x^2+60*x=-80*x-400//разделим на -4
2*x^2-15*x=20*x+100
2*x^2-35*x-100=0
D=1225+8*100=1225+800=2025=>корень из D=45
x1=(35-45)/2*2 - не может
x2=80/4=20 - время 2
x=x2+5=20+5=25

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