1-(2)/(4t-1)=(15)/(4t-1)^2

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Solution for 1-(2)/(4t-1)=(15)/(4t-1)^2 equation:


D( t )

(4*t-1)^2 = 0

4*t-1 = 0

(4*t-1)^2 = 0

(4*t-1)^2 = 0

4*t-1 = 0 // + 1

4*t = 1 // : 4

t = 1/4

4*t-1 = 0

4*t-1 = 0

4*t-1 = 0 // + 1

4*t = 1 // : 4

t = 1/4

t in (-oo:1/4) U (1/4:+oo)

1-(2/(4*t-1)) = 15/((4*t-1)^2) // - 15/((4*t-1)^2)

1-(2/(4*t-1))-(15/((4*t-1)^2)) = 0

1-2*(4*t-1)^-1-15*(4*t-1)^-2 = 0

1-2/(4*t-1)-15/((4*t-1)^2) = 0

(-2*(4*t-1))/((4*t-1)^2)-15/((4*t-1)^2)+(1*(4*t-1)^2)/((4*t-1)^2) = 0

1*(4*t-1)^2-2*(4*t-1)-15 = 0

16*t^2-8*t-8*t-13+1 = 0

16*t^2-16*t-12 = 0

16*t^2-16*t-12 = 0

4*(4*t^2-4*t-3) = 0

4*t^2-4*t-3 = 0

DELTA = (-4)^2-(-3*4*4)

DELTA = 64

DELTA > 0

t = (64^(1/2)+4)/(2*4) or t = (4-64^(1/2))/(2*4)

t = 3/2 or t = -1/2

4*(t+1/2)*(t-3/2) = 0

(4*(t+1/2)*(t-3/2))/((4*t-1)^2) = 0

(4*(t+1/2)*(t-3/2))/((4*t-1)^2) = 0 // * (4*t-1)^2

4*(t+1/2)*(t-3/2) = 0

( 4 )

4 = 0

t belongs to the empty set

( t+1/2 )

t+1/2 = 0 // - 1/2

t = -1/2

( t-3/2 )

t-3/2 = 0 // + 3/2

t = 3/2

t in { -1/2, 3/2 }

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