(-k-3)(6k-1)=0

Solution for (-k-3)(6k-1)=0 equation:

(-k-3)(6k-1)=0

We add all the numbers together, and all the variables

(-1k-3)(6k-1)=0

We multiply parentheses ..

(-6k^2+k-18k+3)=0

We get rid of parentheses

-6k^2+k-18k+3=0

We add all the numbers together, and all the variables

-6k^2-17k+3=0

a = -6; b = -17; c = +3;
Δ = b2-4ac
Δ = -172-4·(-6)·3
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-19}{2*-6}=\frac{-2}{-12}
=1/6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+19}{2*-6}=\frac{36}{-12}
=-3
$