(k+12)(3k-2)=0

Solution for (k+12)(3k-2)=0 equation:

(k+12)(3k-2)=0

We multiply parentheses ..

(+3k^2-2k+36k-24)=0

We get rid of parentheses

3k^2-2k+36k-24=0

We add all the numbers together, and all the variables

3k^2+34k-24=0

a = 3; b = 34; c = -24;
Δ = b2-4ac
Δ = 342-4·3·(-24)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-38}{2*3}=\frac{-72}{6}
=-12
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+38}{2*3}=\frac{4}{6}
=2/3
$