1/3k+4=5/6k-12

Solution for 1/3k+4=5/6k-12 equation:

1/3k+4=5/6k-12

We move all terms to the left:

1/3k+4-(5/6k-12)=0


Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R



Domain of the equation: 6k-12)!=0

k∈R


We get rid of parentheses

1/3k-5/6k+12+4=0

We calculate fractions


6k/18k^2+(-15k)/18k^2+12+4=0

We add all the numbers together, and all the variables

6k/18k^2+(-15k)/18k^2+16=0

We multiply all the terms by the denominator


6k+(-15k)+16*18k^2=0

Wy multiply elements

288k^2+6k+(-15k)=0

We get rid of parentheses

288k^2+6k-15k=0

We add all the numbers together, and all the variables

288k^2-9k=0

a = 288; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·288·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*288}=\frac{0}{576}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*288}=\frac{18}{576}
=1/32
$