6(k+12)3k=3

Solution for 6(k+12)3k=3 equation:

6(k+12)3k=3

We move all terms to the left:

6(k+12)3k-(3)=0

We multiply parentheses

18k^2+216k-3=0

a = 18; b = 216; c = -3;
Δ = b2-4ac
Δ = 2162-4·18·(-3)
Δ = 46872
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:

$\sqrt{\Delta}=\sqrt{46872}=\sqrt{36*1302}=\sqrt{36}*\sqrt{1302}=6\sqrt{1302}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(216)-6\sqrt{1302}}{2*18}=\frac{-216-6\sqrt{1302}}{36}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(216)+6\sqrt{1302}}{2*18}=\frac{-216+6\sqrt{1302}}{36}
$