(x+7)/5+(x-5)/3=x2

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Solution for (x+7)/5+(x-5)/3=x2 equation: 


x in (-oo:+oo)

(x+7)/5+(x-5)/3 = x^2 // - x^2

(x+7)/5+(x-5)/3-x^2 = 0

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

3*(x+7)+5*(x-5)+3*5*(-x^2) = 0

8*x-15*x^2-4 = 0

8*x-15*x^2-4 = 0

8*x-15*x^2-4 = 0

DELTA = 8^2-(-15*(-4)*4)

DELTA = -176

DELTA < 0

1 = 0

1/(3*5) = 0

1/(3*5) = 0 // * 3*5

1 = 0

x belongs to the empty set

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