1/3k+80=k+120

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

1/3k+80=k+120

We move all terms to the left:

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


Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


-k*3k-120*3k+80*3k+1=0

Wy multiply elements

-3k^2-360k+240k+1=0

We add all the numbers together, and all the variables

-3k^2-120k+1=0

a = -3; b = -120; c = +1;
Δ = b2-4ac
Δ = -1202-4·(-3)·1
Δ = 14412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14412}=\sqrt{4*3603}=\sqrt{4}*\sqrt{3603}=2\sqrt{3603}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-2\sqrt{3603}}{2*-3}=\frac{120-2\sqrt{3603}}{-6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+2\sqrt{3603}}{2*-3}=\frac{120+2\sqrt{3603}}{-6}
$