20=(n+5)(n-2)

Solution for 20=(n+5)(n-2) equation:

20=(n+5)(n-2)

We move all terms to the left:

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

We multiply parentheses ..

-((+n^2-2n+5n-10))+20=0


We calculate terms in parentheses: -((+n^2-2n+5n-10)), so:

(+n^2-2n+5n-10)

We get rid of parentheses

n^2-2n+5n-10

We add all the numbers together, and all the variables

n^2+3n-10

Back to the equation:

-(n^2+3n-10)


We get rid of parentheses

-n^2-3n+10+20=0

We add all the numbers together, and all the variables

-1n^2-3n+30=0

a = -1; b = -3; c = +30;
Δ = b2-4ac
Δ = -32-4·(-1)·30
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{129}}{2*-1}=\frac{3-\sqrt{129}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*-1}=\frac{3+\sqrt{129}}{-2}
$