50=(n-4)(n+5)

Solution for 50=(n-4)(n+5) equation:

50=(n-4)(n+5)

We move all terms to the left:

50-((n-4)(n+5))=0

We multiply parentheses ..

-((+n^2+5n-4n-20))+50=0


We calculate terms in parentheses: -((+n^2+5n-4n-20)), so:

(+n^2+5n-4n-20)

We get rid of parentheses

n^2+5n-4n-20

We add all the numbers together, and all the variables

n^2+n-20

Back to the equation:

-(n^2+n-20)


We get rid of parentheses

-n^2-n+20+50=0

We add all the numbers together, and all the variables

-1n^2-1n+70=0

a = -1; b = -1; c = +70;
Δ = b2-4ac
Δ = -12-4·(-1)·70
Δ = 281
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{-(-1)-\sqrt{281}}{2*-1}=\frac{1-\sqrt{281}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{281}}{2*-1}=\frac{1+\sqrt{281}}{-2}
$