12k(k+2)=5k-4

Solution for 12k(k+2)=5k-4 equation:

12k(k+2)=5k-4

We move all terms to the left:

12k(k+2)-(5k-4)=0

We multiply parentheses

12k^2+24k-(5k-4)=0

We get rid of parentheses

12k^2+24k-5k+4=0

We add all the numbers together, and all the variables

12k^2+19k+4=0

a = 12; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·12·4
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*12}=\frac{-32}{24}
=-1+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*12}=\frac{-6}{24}
=-1/4
$