16x^2-(1/2)x+(1/16)=x-4

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Solution for 16x^2-(1/2)x+(1/16)=x-4 equation: 


x in (-oo:+oo)

16*x^2-((1/2)*x)+1/16 = x-4 // - x-4

16*x^2-((1/2)*x)-x+1/16+4 = 0

16*x^2+(-1/2)*x-x+1/16+4 = 0

16*x^2-3/2*x+65/16 = 0

DELTA = (-3/2)^2-(4*65/16*16)

DELTA = -1031/4

DELTA < 0

x belongs to the empty set

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