20n(n+2)=9n-12

Solution for 20n(n+2)=9n-12 equation:

20n(n+2)=9n-12

We move all terms to the left:

20n(n+2)-(9n-12)=0

We multiply parentheses

20n^2+40n-(9n-12)=0

We get rid of parentheses

20n^2+40n-9n+12=0

We add all the numbers together, and all the variables

20n^2+31n+12=0

a = 20; b = 31; c = +12;
Δ = b2-4ac
Δ = 312-4·20·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*20}=\frac{-32}{40}
=-4/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*20}=\frac{-30}{40}
=-3/4
$