(x^2+x)/2=(3x+1)/9

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Solution for (x^2+x)/2=(3x+1)/9 equation: 


x in (-oo:+oo)

(x^2+x)/2 = (3*x+1)/9 // - (3*x+1)/9

(x^2+x)/2-((3*x+1)/9) = 0

(x^2+x)/2+(-1*(3*x+1))/9 = 0

(9*(x^2+x))/(2*9)+(-1*2*(3*x+1))/(2*9) = 0

9*(x^2+x)-1*2*(3*x+1) = 0

9*x^2+3*x-2 = 0

9*x^2+3*x-2 = 0

9*x^2+3*x-2 = 0

DELTA = 3^2-(-2*4*9)

DELTA = 81

DELTA > 0

x = (81^(1/2)-3)/(2*9) or x = (-81^(1/2)-3)/(2*9)

x = 1/3 or x = -2/3

(x+2/3)*(x-1/3) = 0

((x+2/3)*(x-1/3))/(2*9) = 0

((x+2/3)*(x-1/3))/(2*9) = 0 // * 2*9

(x+2/3)*(x-1/3) = 0

( x+2/3 )

x+2/3 = 0 // - 2/3

x = -2/3

( x-1/3 )

x-1/3 = 0 // + 1/3

x = 1/3

x in { -2/3, 1/3 }

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