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

Solution for (3k-10)(k+12)=0 equation:

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

We multiply parentheses ..

(+3k^2+36k-10k-120)=0

We get rid of parentheses

3k^2+36k-10k-120=0

We add all the numbers together, and all the variables

3k^2+26k-120=0

a = 3; b = 26; c = -120;
Δ = b2-4ac
Δ = 262-4·3·(-120)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-46}{2*3}=\frac{-72}{6}
=-12
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+46}{2*3}=\frac{20}{6}
=3+1/3
$