N=(x^2-3x)/5

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Solution for N=(x^2-3x)/5 equation: 


x in (-oo:+oo)

N = (x^2-(3*x))/5 // - (x^2-(3*x))/5

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

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

(-1*(x^2-3*x))/5+N = 0

(-1*(x^2-3*x))/5+(5*N)/5 = 0

5*N-1*(x^2-3*x) = 0

5*N-x^2+3*x = 0

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

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

5*N-x^2+3*x = 0

5*N-x^2+3*x = 0

DELTA = 3^2-(-1*4*5*N)

DELTA = 20*N+9

20*N+9 = 0

20*N+9 = 0 // - 9

20*N = -9 // : 20

N = -9/20

DELTA = 0  <=>  t_3 = -9/20

x = -3/(-1*2) i N = -9/20

x = 3/2 i N = -9/20

( x = ((20*N+9)^(1/2)-3)/(-1*2) or x = (-(20*N+9)^(1/2)-3)/(-1*2) ) i N > -9/20

( x = ((20*N+9)^(1/2)-3)/(-2) or x = ((20*N+9)^(1/2)+3)/2 ) i N > -9/20

N-(-9/20) > 0

N+9/20 > 0

N+9/20 > 0 // - 9/20

N > -9/20

x in { 3/2, ((20*N+9)^(1/2)-3)/(-2), ((20*N+9)^(1/2)+3)/2 }

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