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Объяснение:

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Пусть собственная скорость лодки x [км/ч], тогда скорость лодки по течению x+2 [км/ч] и против течения x-2 [км/ч]. 
Время, затраченное на первый отрезок пути: 16/(x-2) [ч], 
на второй отрезок пути: 12/(x+2) [ч]. 
Общее время в пути: 16/(x-2) + 12/(x+2) = 3 [ч] 
x <>2 и x <> -2, домножаем обе части уравнения на (x+2)*(x-2), получаем: 
16*(x+2) + 12*(x-2) = 3*(x+2)*(x-2) 
16*x + 32 + 12*x - 24 = 3* x^2 - 12, где x^2 = x*x 
28*x + 8 = 3* x^2 - 12 
3*x^2 - 28*x - 20 = 0 
Дискриминант: D = b^2 - 4*a*c = 28*28 - 4*3*(-20) = 1024 = 32^2 
x1 = (-b + sqrt(D))/(2*a) = (28 + 32) / 6 = 10 [км/ч] 
x2 = (-b - sqrt(D))/(2*a) = (28 - 32) / 6 = -2/3 [км/ч] 
Второй корень логически не имеет смысла, поэтому ответ: 10 км/ч.