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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(k(3k+1)), so:

k(3k+1)

We multiply parentheses

3k^2+k

Back to the equation:

-(3k^2+k)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-3k^2+5k+2=0

a = -3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·(-3)·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-3}=\frac{-12}{-6}
=+2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-3}=\frac{2}{-6}
=-1/3
$