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

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Solution for 4x^2(5+x)/20x+4x^2=0 equation: 


x in (-oo:+oo)

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

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

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

4*x*x^2*(x+5)+4*20*x^2 = 0

4*x^4+20*x^3+80*x^2 = 0

4*x^4+20*x^3+80*x^2 = 0

4*x^2*(x^2+5*x+20) = 0

x^2+5*x+20 = 0

DELTA = 5^2-(1*4*20)

DELTA = -55

DELTA < 0

4*x^2 = 0

(4*x^2)/20 = 0

(4*x^2)/20 = 0 // * 20

4*x^2 = 0

4*x^2 = 0 // : 4

x^2 = 0

x = 0

x = 0

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