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

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Solution for (n-8)(n-2)=0 equation:



(n-8)(n-2)=0
We multiply parentheses ..
(+n^2-2n-8n+16)=0
We get rid of parentheses
n^2-2n-8n+16=0
We add all the numbers together, and all the variables
n^2-10n+16=0
a = 1; b = -10; c = +16;
Δ = b2-4ac
Δ = -102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*1}=\frac{4}{2} =2 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*1}=\frac{16}{2} =8 $

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