2(3k+1)k=18

Solution for 2(3k+1)k=18 equation:

2(3k+1)k=18

We move all terms to the left:

2(3k+1)k-(18)=0

We multiply parentheses

6k^2+2k-18=0

a = 6; b = 2; c = -18;
Δ = b2-4ac
Δ = 22-4·6·(-18)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{109}}{2*6}=\frac{-2-2\sqrt{109}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{109}}{2*6}=\frac{-2+2\sqrt{109}}{12}
$