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

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

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

We multiply parentheses ..

(+3k^2-1k+3k-1)=0

We get rid of parentheses

3k^2-1k+3k-1=0

We add all the numbers together, and all the variables

3k^2+2k-1=0

a = 3; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·3·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*3}=\frac{-6}{6}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*3}=\frac{2}{6}
=1/3
$