(3/x-4)+(5/x)=-20/x^2-4

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Solution for (3/x-4)+(5/x)=-20/x^2-4 equation: 


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

x in (-oo:0) U (0:+oo)

3/x+5/x-4 = -20/(x^2)-4 // + -20/(x^2)-4

3/x+5/x-(-20/(x^2))-4+4 = 0

3/x+5/x+20*x^-2-4+4 = 0

8*x^-1+20*x^-2 = 0

t_1 = x^-1

20*t_1^2+8*t_1^1 = 0

20*t_1^2+8*t_1 = 0

DELTA = 8^2-(0*4*20)

DELTA = 64

DELTA > 0

t_1 = (64^(1/2)-8)/(2*20) or t_1 = (-64^(1/2)-8)/(2*20)

t_1 = 0 or t_1 = -2/5

t_1 = -2/5

x^-1+2/5 = 0

1*x^-1 = -2/5 // : 1

x^-1 = -2/5

-1 < 0

1/(x^1) = -2/5 // * x^1

1 = -2/5*x^1 // : -2/5

-5/2 = x^1

x = -5/2

t_1 = 0

x^-1+0 = 0

x^-1 = 0

1*x^-1 = 0 // : 1

x^-1 = 0

x naleu017Cy do O

x = -5/2

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