1/3x^2-4x-Y=0

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Solution for 1/3x^2-4x-Y=0 equation: 


x in (-oo:+oo)

(1/3)*x^2-Y-(4*x) = 0

(1/3)*x^2-Y-4*x = 0

1/3*x^2-Y-4*x = 0

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

DELTA = 4/3*Y+16

4/3*Y+16 = 0

4/3*Y+16 = 0 // - 16

4/3*Y = -16 // : 4/3

Y = -16/4/3

Y = -12

DELTA = 0  <=>  t_1 = -12

x = 4/(1/3*2) i Y = -12

x = 6 i Y = -12

( x = ((4/3*Y+16)^(1/2)+4)/(1/3*2) or x = (4-(4/3*Y+16)^(1/2))/(1/3*2) ) i Y > -12

( x = 3/2*((4/3*Y+16)^(1/2)+4) or x = 3/2*(4-(4/3*Y+16)^(1/2)) ) i Y > -12

Y+12 > 0

Y+12 > 0 // - 12

Y > -12

x in { 6, 3/2*((4/3*Y+16)^(1/2)+4), 3/2*(4-(4/3*Y+16)^(1/2)) }

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