2x^4/3-5x^2/3+2=0

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Solution for 2x^4/3-5x^2/3+2=0 equation: 


x in (-oo:+oo)

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

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

2/3*x^4-5/3*x^2+2 = 0

t_1 = x^2

2/3*t_1^2-5/3*t_1^1+2 = 0

2/3*t_1^2-5/3*t_1+2 = 0

DELTA = (-5/3)^2-(2/3*2*4)

DELTA = -23/9

DELTA < 0

x belongs to the empty set

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