3k(k-1)=2(k+6)

Solution for 3k(k-1)=2(k+6) equation:

3k(k-1)=2(k+6)

We move all terms to the left:

3k(k-1)-(2(k+6))=0

We multiply parentheses

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


We calculate terms in parentheses: -(2(k+6)), so:

2(k+6)

We multiply parentheses

2k+12

Back to the equation:

-(2k+12)


We get rid of parentheses

3k^2-3k-2k-12=0

We add all the numbers together, and all the variables

3k^2-5k-12=0

a = 3; b = -5; c = -12;
Δ = b2-4ac
Δ = -52-4·3·(-12)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*3}=\frac{-8}{6}
=-1+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*3}=\frac{18}{6}
=3
$