(2k-4)(k+3)=0

Solution for (2k-4)(k+3)=0 equation:

(2k-4)(k+3)=0

We multiply parentheses ..

(+2k^2+6k-4k-12)=0

We get rid of parentheses

2k^2+6k-4k-12=0

We add all the numbers together, and all the variables

2k^2+2k-12=0

a = 2; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·2·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*2}=\frac{-12}{4}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*2}=\frac{8}{4}
=2
$