(2x^2-9x-5)/(4x^2+2x^2)=0

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Solution for (2x^2-9x-5)/(4x^2+2x^2)=0 equation: 


D( x )

4*x^2+2*x^2 = 0

4*x^2+2*x^2 = 0

4*x^2+2*x^2 = 0

6*x^2 = 0 // : 6

x^2 = 0

x = 0

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

(2*x^2-(9*x)-5)/(4*x^2+2*x^2) = 0

(2*x^2-9*x-5)/(4*x^2+2*x^2) = 0

(2*x^2-9*x-5)/(6*x^2) = 0

2*x^2-9*x-5 = 0

2*x^2-9*x-5 = 0

DELTA = (-9)^2-(-5*2*4)

DELTA = 121

DELTA > 0

x = (121^(1/2)+9)/(2*2) or x = (9-121^(1/2))/(2*2)

x = 5 or x = -1/2

(x+1/2)*(x-5) = 0

((x+1/2)*(x-5))/(6*x^2) = 0

( x+1/2 )

x+1/2 = 0 // - 1/2

x = -1/2

( x-5 )

x-5 = 0 // + 5

x = 5

x in { -1/2, 5 }

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