20n(n+3)=19n-20

Solution for 20n(n+3)=19n-20 equation:

20n(n+3)=19n-20

We move all terms to the left:

20n(n+3)-(19n-20)=0

We multiply parentheses

20n^2+60n-(19n-20)=0

We get rid of parentheses

20n^2+60n-19n+20=0

We add all the numbers together, and all the variables

20n^2+41n+20=0

a = 20; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·20·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-9}{2*20}=\frac{-50}{40}
=-1+1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+9}{2*20}=\frac{-32}{40}
=-4/5
$