x^2+4+(36-20x+16x^2)/3

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Solution for x^2+4+(36-20x+16x^2)/3 equation: 


x in (-oo:+oo)

(16*x^2-(20*x)+36)/3+x^2+4 = 0

(16*x^2-20*x+36)/3+x^2+4 = 0

(16*x^2-20*x+36)/3+(3*x^2)/3+(3*4)/3 = 0

16*x^2+3*x^2-20*x+3*4+36 = 0

19*x^2-20*x+12+36 = 0

19*x^2-20*x+48 = 0

19*x^2-20*x+48 = 0

19*x^2-20*x+48 = 0

DELTA = (-20)^2-(4*19*48)

DELTA = -3248

DELTA < 0

1 = 0

1/3 = 0

1/3 = 0 // * 3

1 = 0

x belongs to the empty set

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