-(1/2)x^2+x-2=y

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


x in (-oo:+oo)

x-((1/2)*x^2)-2 = y // - y

x-((1/2)*x^2)-y-2 = 0

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

x-1/2*x^2-y-2 = 0

DELTA = 1^2-(-1/2*4*(-y-2))

DELTA = 1-2*(y+2)

1-2*(y+2) = 0

1-2*(y+2) = 0

-2*y-3 = 0

-2*y-3 = 0

-2*y-3 = 0 // + 3

-2*y = 3 // : -2

y = 3/(-2)

y = -3/2

DELTA = 0  <=>  t_1 = -3/2

x = -1/(-1/2*2) i y = -3/2

x = 1 i y = -3/2

( x = ((1-2*(y+2))^(1/2)-1)/(-1/2*2) or x = (-(1-2*(y+2))^(1/2)-1)/(-1/2*2) ) i y > -3/2

( x = 1-(1-2*(y+2))^(1/2) or x = (1-2*(y+2))^(1/2)+1 ) i y > -3/2

y+3/2 > 0

y+3/2 > 0 // - 3/2

y > -3/2

x in { 1, 1-(1-2*(y+2))^(1/2), (1-2*(y+2))^(1/2)+1 }

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