(1)/(3)k+80=(1)/(2)k+120

Solution for (1)/(3)k+80=(1)/(2)k+120 equation:

(1)/(3)k+80=(1)/(2)k+120

We move all terms to the left:

(1)/(3)k+80-((1)/(2)k+120)=0


Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R



Domain of the equation: 2k+120)!=0

k∈R


We get rid of parentheses

1/3k-1/2k-120+80=0

We calculate fractions


2k/6k^2+(-3k)/6k^2-120+80=0

We add all the numbers together, and all the variables

2k/6k^2+(-3k)/6k^2-40=0

We multiply all the terms by the denominator


2k+(-3k)-40*6k^2=0

Wy multiply elements

-240k^2+2k+(-3k)=0

We get rid of parentheses

-240k^2+2k-3k=0

We add all the numbers together, and all the variables

-240k^2-1k=0

a = -240; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-240)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-240}=\frac{0}{-480}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-240}=\frac{2}{-480}
=-1/240
$