(y-2/5)^2=-7

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Solution for (y-2/5)^2=-7 equation: 


y in (-oo:+oo)

(y-(2/5))^2 = -7 // + 7

(y-(2/5))^2+7 = 0

(y-2/5)^2+7 = 0

(y-2/5)^2+7 = 0

y^2-4/5*y+179/25 = 0

y^2-4/5*y+179/25 = 0

y^2-4/5*y+179/25 = 0

DELTA = (-4/5)^2-(1*4*179/25)

DELTA = -28

DELTA < 0

1 = 0

1 = 0

y belongs to the empty set

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