(n-8)(2n-2=5)(n-8)

Solution for (n-8)(2n-2=5)(n-8) equation:

(n-8)(2n-2=5)(n-8)

We move all terms to the left:

(n-8)(2n-2-(5)(n-8))=0


We calculate terms in parentheses: +(n-8)(2n-2-5(n-8)), so:

n-8)(2n-2-5(n-8)

determiningTheFunctionDomain
n-8)(2n-5(n-8)-2

We multiply parentheses

n-8)(2n-5n+40-2

We add all the numbers together, and all the variables

-4n-8)(2n+38

Back to the equation:

+(-4n-8)(2n+38)


We multiply parentheses ..

(-8n^2-152n-16n-304)=0

We get rid of parentheses

-8n^2-152n-16n-304=0

We add all the numbers together, and all the variables

-8n^2-168n-304=0

a = -8; b = -168; c = -304;
Δ = b2-4ac
Δ = -1682-4·(-8)·(-304)
Δ = 18496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{18496}=136$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-168)-136}{2*-8}=\frac{32}{-16}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-168)+136}{2*-8}=\frac{304}{-16}
=-19
$