N-5=5(n-8)(n-10)

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Solution for N-5=5(n-8)(n-10) equation: 



-5=5(N-8)(N-10)

We move all terms to the left:

-5-(5(N-8)(N-10))=0

We multiply parentheses ..

-(5(+N^2-10N-8N+80))-5=0



We calculate terms in parentheses: -(5(+N^2-10N-8N+80)), so:

5(+N^2-10N-8N+80)

We multiply parentheses

5N^2-50N-40N+400

We add all the numbers together, and all the variables

5N^2-90N+400

Back to the equation:

-(5N^2-90N+400)



We get rid of parentheses

-5N^2+90N-400-5=0

We add all the numbers together, and all the variables

-5N^2+90N-405=0

a = -5; b = 90; c = -405;
Δ = b2-4ac
Δ = 902-4·(-5)·(-405)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$N=\frac{-b}{2a}=\frac{-90}{-10}=+9$

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