12k(k+3)=11k-12

Solution for 12k(k+3)=11k-12 equation:

12k(k+3)=11k-12

We move all terms to the left:

12k(k+3)-(11k-12)=0

We multiply parentheses

12k^2+36k-(11k-12)=0

We get rid of parentheses

12k^2+36k-11k+12=0

We add all the numbers together, and all the variables

12k^2+25k+12=0

a = 12; b = 25; c = +12;
Δ = b2-4ac
Δ = 252-4·12·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-7}{2*12}=\frac{-32}{24}
=-1+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+7}{2*12}=\frac{-18}{24}
=-3/4
$