2(k-5)3k=k+6

Solution for 2(k-5)3k=k+6 equation:

2(k-5)3k=k+6

We move all terms to the left:

2(k-5)3k-(k+6)=0

We multiply parentheses

6k^2-30k-(k+6)=0

We get rid of parentheses

6k^2-30k-k-6=0

We add all the numbers together, and all the variables

6k^2-31k-6=0

a = 6; b = -31; c = -6;
Δ = b2-4ac
Δ = -312-4·6·(-6)
Δ = 1105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{1105}}{2*6}=\frac{31-\sqrt{1105}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1105}}{2*6}=\frac{31+\sqrt{1105}}{12}
$