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

Solution for (3k+2)(k-1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3k^2-1k-2=0

a = 3; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·3·(-2)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*3}=\frac{-4}{6}
=-2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*3}=\frac{6}{6}
=1
$