(x+4)^2/5-7(x+4)^1/5+12=0

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Solution for (x+4)^2/5-7(x+4)^1/5+12=0 equation: 


x in (-oo:+oo)

((x+4)^2)/5-((7*(x+4)^1)/5)+12 = 0

((x+4)^2)/5+(-7/5)*(x+4)+12 = 0

((x+4)^2)/5-7/5*(x+4)+12 = 0

((x+4)^2)/5+(-7/5*5*(x+4))/5+(5*12)/5 = 0

(x+4)^2-7/5*5*(x+4)+5*12 = 0

x^2+x-12+60 = 0

x^2+x+48 = 0

x^2+x+48 = 0

x^2+x+48 = 0

DELTA = 1^2-(1*4*48)

DELTA = -191

DELTA < 0

1 = 0

1/5 = 0

1/5 = 0 // * 5

1 = 0

x belongs to the empty set

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