(1/5)(2z-5)^2=(2/3)-2

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Solution for (1/5)(2z-5)^2=(2/3)-2 equation: 


z in (-oo:+oo)

(1/5)*(2*z-5)^2 = 2/3-2 // - 2/3-2

(1/5)*(2*z-5)^2-(2/3)+2 = 0

(1/5)*(2*z-5)^2-2/3+2 = 0

1/5*(2*z-5)^2-2/3+2 = 0

(1/5*3*(2*z-5)^2)/3-2/3+(2*3)/3 = 0

1/5*3*(2*z-5)^2+2*3-2 = 0

12/5*z^2-12*z+6+13 = 0

12/5*z^2-12*z+19 = 0

12/5*z^2-12*z+19 = 0

12/5*z^2-12*z+19 = 0

DELTA = (-12)^2-(12/5*4*19)

DELTA = -192/5

DELTA < 0

1 = 0

1/3 = 0

1/3 = 0 // * 3

1 = 0

1 = 0

z belongs to the empty set

z belongs to the empty set

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