12n(n+2)=n-5

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

12n(n+2)=n-5

We move all terms to the left:

12n(n+2)-(n-5)=0

We multiply parentheses

12n^2+24n-(n-5)=0

We get rid of parentheses

12n^2+24n-n+5=0

We add all the numbers together, and all the variables

12n^2+23n+5=0

a = 12; b = 23; c = +5;
Δ = b2-4ac
Δ = 232-4·12·5
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*12}=\frac{-40}{24}
=-1+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*12}=\frac{-6}{24}
=-1/4
$