(6k+12)(4k-7)=0

Solution for (6k+12)(4k-7)=0 equation:

(6k+12)(4k-7)=0

We multiply parentheses ..

(+24k^2-42k+48k-84)=0

We get rid of parentheses

24k^2-42k+48k-84=0

We add all the numbers together, and all the variables

24k^2+6k-84=0

a = 24; b = 6; c = -84;
Δ = b2-4ac
Δ = 62-4·24·(-84)
Δ = 8100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8100}=90$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-90}{2*24}=\frac{-96}{48}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+90}{2*24}=\frac{84}{48}
=1+3/4
$