2n+3(n-10)n=20

Solution for 2n+3(n-10)n=20 equation:

2n+3(n-10)n=20

We move all terms to the left:

2n+3(n-10)n-(20)=0

We multiply parentheses

3n^2+2n-30n-20=0

We add all the numbers together, and all the variables

3n^2-28n-20=0

a = 3; b = -28; c = -20;
Δ = b2-4ac
Δ = -282-4·3·(-20)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-32}{2*3}=\frac{-4}{6}
=-2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+32}{2*3}=\frac{60}{6}
=10
$