((3^3-10)+(-7))/((-5)^2+(-4x-5))

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Solution for ((3^3-10)+(-7))/((-5)^2+(-4x-5)) equation: 


D( x )

(-5)^2-4*x-5 = 0

(-5)^2-4*x-5 = 0

(-5)^2-4*x-5 = 0

20-4*x = 0 // - 20

-4*x = -20 // : -4

x = -20/(-4)

x = 5

x in (-oo:5) U (5:+oo)

(3^3-10-7)/((-5)^2-4*x-5) = 0

10/(20-4*x) = 0

x belongs to the empty set

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