12/(k+3)+2=-15/(k-4)

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Solution for 12/(k+3)+2=-15/(k-4) equation:


D( k )

k-4 = 0

k+3 = 0

k-4 = 0

k-4 = 0

k-4 = 0 // + 4

k = 4

k+3 = 0

k+3 = 0

k+3 = 0 // - 3

k = -3

k in (-oo:-3) U (-3:4) U (4:+oo)

12/(k+3)+2 = -15/(k-4) // + -15/(k-4)

12/(k+3)-(-15/(k-4))+2 = 0

12/(k+3)+15*(k-4)^-1+2 = 0

12/(k+3)+15/(k-4)+2 = 0

(12*(k-4))/((k+3)*(k-4))+(15*(k+3))/((k+3)*(k-4))+(2*(k+3)*(k-4))/((k+3)*(k-4)) = 0

12*(k-4)+15*(k+3)+2*(k+3)*(k-4) = 0

2*k^2+27*k-2*k-24-3 = 0

2*k^2+25*k-27 = 0

2*k^2+25*k-27 = 0

2*k^2+25*k-27 = 0

DELTA = 25^2-(-27*2*4)

DELTA = 841

DELTA > 0

k = (841^(1/2)-25)/(2*2) or k = (-841^(1/2)-25)/(2*2)

k = 1 or k = -27/2

(k+27/2)*(k-1) = 0

((k+27/2)*(k-1))/((k+3)*(k-4)) = 0

((k+27/2)*(k-1))/((k+3)*(k-4)) = 0 // * (k+3)*(k-4)

(k+27/2)*(k-1) = 0

( k+27/2 )

k+27/2 = 0 // - 27/2

k = -27/2

( k-1 )

k-1 = 0 // + 1

k = 1

k in { -27/2, 1 }

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