x^2-3x+y^2+2x=3/4

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Solution for x^2-3x+y^2+2x=3/4 equation:


x in (-oo:+oo)

t_1 = y^2-3/4

t_1+x^2-3*x+2*x = 0

t_1+x^2-x = 0

DELTA = (-1)^2-(1*4*t_1)

DELTA = 1-4*t_1

1-4*t_1 = 0

1-4*t_1 = 0 // - 1

-4*t_1 = -1 // : -4

t_1 = -1/(-4)

t_1 = 1/4

DELTA = 0 <=> t_2 = 1/4

x = 1/(1*2) i t_1 = 1/4

x = 1/2 i t_1 = 1/4

( x = ((1-4*t_1)^(1/2)+1)/(1*2) or x = (1-(1-4*t_1)^(1/2))/(1*2) ) i t_1 > 1/4

( x = ((1-4*t_1)^(1/2)+1)/2 or x = (1-(1-4*t_1)^(1/2))/2 ) i t_1 > 1/4

t_1-1/4 > 0

t_1-1/4 > 0 // + 1/4

t_1 > 1/4

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

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