15k(k+2)=k-12

Solution for 15k(k+2)=k-12 equation:

15k(k+2)=k-12

We move all terms to the left:

15k(k+2)-(k-12)=0

We multiply parentheses

15k^2+30k-(k-12)=0

We get rid of parentheses

15k^2+30k-k+12=0

We add all the numbers together, and all the variables

15k^2+29k+12=0

a = 15; b = 29; c = +12;
Δ = b2-4ac
Δ = 292-4·15·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-11}{2*15}=\frac{-40}{30}
=-1+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+11}{2*15}=\frac{-18}{30}
=-3/5
$