решите уравнение : ___________ 3x^2+7x-40=0
(^ - знак квадрата)
3x^+7x-40 = 0
D = 7^ -4*3*(-40) = 49+480 = 529
x1 = (-7+23)/3*2 = 16/6 = 8/3 = 2 2/3
x2 = (-7-23)/2*3 = -30/6 = -5
Ответ: 2 2/3; -5.
D=49+4*3*40=549=sqrt(549)=23
x1=-7-23/6=-5
x2=-7+23/6=сам посчитаешь))
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(^ - знак квадрата)
3x^+7x-40 = 0
D = 7^ -4*3*(-40) = 49+480 = 529
x1 = (-7+23)/3*2 = 16/6 = 8/3 = 2 2/3
x2 = (-7-23)/2*3 = -30/6 = -5
Ответ: 2 2/3; -5.
D=49+4*3*40=549=sqrt(549)=23
x1=-7-23/6=-5
x2=-7+23/6=сам посчитаешь))