k(3k-6)-7k(6k+10)=12

Solution for k(3k-6)-7k(6k+10)=12 equation:

k(3k-6)-7k(6k+10)=12

We move all terms to the left:

k(3k-6)-7k(6k+10)-(12)=0

We multiply parentheses

3k^2-42k^2-6k-70k-12=0

We add all the numbers together, and all the variables

-39k^2-76k-12=0

a = -39; b = -76; c = -12;
Δ = b2-4ac
Δ = -762-4·(-39)·(-12)
Δ = 3904
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:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-76)-8\sqrt{61}}{2*-39}=\frac{76-8\sqrt{61}}{-78}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-76)+8\sqrt{61}}{2*-39}=\frac{76+8\sqrt{61}}{-78}
$