3/(h+4)-4=6/(h^2+h-12)

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Solution for 3/(h+4)-4=6/(h^2+h-12) equation: 


D( h )

h+4 = 0

h^2+h-12 = 0

h+4 = 0

h+4 = 0

h+4 = 0 // - 4

h = -4

h^2+h-12 = 0

h^2+h-12 = 0

h^2+h-12 = 0

DELTA = 1^2-(-12*1*4)

DELTA = 49

DELTA > 0

h = (49^(1/2)-1)/(1*2) or h = (-49^(1/2)-1)/(1*2)

h = 3 or h = -4

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

3/(h+4)-4 = 6/(h^2+h-12) // - 6/(h^2+h-12)

3/(h+4)-(6/(h^2+h-12))-4 = 0

3/(h+4)-6*(h^2+h-12)^-1-4 = 0

3/(h+4)-6/(h^2+h-12)-4 = 0

h^2+h-12 = 0

h^2+h-12 = 0

h^2+h-12 = 0

DELTA = 1^2-(-12*1*4)

DELTA = 49

DELTA > 0

h = (49^(1/2)-1)/(1*2) or h = (-49^(1/2)-1)/(1*2)

h = 3 or h = -4

(h+4)*(h-3) = 0

3/(h+4)-6/((h+4)*(h-3))-4 = 0

3/(h+4)-6/(h+4)+(-4*(h+4))/(h+4) = 0

3-4*(h+4)-6 = 0

-4*h-16-3 = 0

-4*h-19 = 0

(-4*h-19)/(h+4) = 0

(-4*h-19)/(h+4) = 0 // * h+4

-4*h-19 = 0

-4*h-19 = 0 // + 19

-4*h = 19 // : -4

h = 19/(-4)

h = -19/4

h = -19/4

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