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

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


x in (-oo:+oo)

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

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

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

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

DELTA = -31/4

DELTA < 0

x belongs to the empty set

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